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Finding Modular Inverses

Cathy Frey · YouTube · 5:00

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This 120-second whiteboard lecture introduces modular inverses. It first defines multiplicative inverses modulo m by the condition ab ≡ 1 (mod m), then states the theorem that an integer a has an inverse modulo m if and only if gcd(a,m)=1. The remainder of the clip works three concrete examples—finding the inverses of 5 mod 7, 5 mod 11, and 6 mod 17—using a repeated search method: start at 1, keep adding the modulus until a multiple of the target integer appears, then factor that multiple to read off the inverse. This 120-second whiteboard clip teaches how to find modular inverses modulo 19. It first solves 'inverse of 18 mod 19' by a slow brute-force congruence chain, then shows a faster shortcut using negative residues: 18 ≡ -1, so 18 is its own inverse. A second example finds the inverse of 17 mod 19 by rewriting 17 as -2, using (-2)^{-1} ≡ -10, and converting -10 to the positive residue 9. This 60-second whiteboard clip finishes two modular-inverse examples modulo 19 and then introduces a nonexample modulo 10. First, the board shows 18≡−1(mod19)18 \equiv -1 \pmod{19} leading to 18−1≡1(mod19)18^{-1} \equiv 1 \pmod{19}, and 17≡−2(mod19)17 \equiv -2 \pmod{19} with (-2)^{-1} ≡−10(mod19)\equiv -10 \pmod{19}, giving 17−1≡9(mod19)17^{-1} \equiv 9 \pmod{19}. The audio adds that a longer search would also reach 17×9=15317 \times 9 = 153. The screen then switches to "Find the inverse of 5 mod 10." The presenter begins the direct search 1≡11≡21≡31≡⋯1 \equiv 11 \equiv 21 \equiv 31 \equiv \cdots, explains that adding 10 never produces a multiple of 5, and then writes (5,10)=5 NOT 1 followed by No inverse. The spoken takeaway is that if the greatest common divisor of the two values is greater than 1, then there is no modular inverse.

Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.

Chapters

0:00Definition of multiplicative inverses modulo m0:23Theorem: inverse exists iff gcd(a,m)=10:40Example: inverse of 5 mod 71:13Example: inverse of 5 mod 111:44Example: inverse of 6 mod 172:00New problem: inverse of 18 mod 192:08Brute-force congruence chain2:43Shortcut using 18 ≡ -13:14Second example: inverse of 17 mod 193:43Convert -10 to positive residue 94:00Completed inverses modulo 194:14Switch to inverse of 5 modulo 104:20Attempted search by adding 104:40gcd obstruction and no inverse4:59Closing remarks

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

The clip opens on a whiteboard-style slide titled “Modular Inverses.” The first mathematical content is a definition: integers a and b are called multiplicative inverses modulo m exactly when ab ≡ 1 (mod m). The speaker restates this informally as “two numbers that multiply together to get you one mod m,” making clear that the inverse is defined through a congruence condition rather than ordinary real-number division.

The slide then presents the central theorem of the segment: the integer a has an inverse modulo m if and only if (a,m)=1. Here (a,m) is read aloud as the greatest common divisor of a and m. The speaker emphasizes the converse direction in words as well: if gcd(a,m)>1, then no inverse modulo m exists. This establishes the existence criterion that governs the examples that follow.

The lecture moves to a worked example: “Find the inverse of 5 mod 7.” The method shown is to search within the congruence class of 1 modulo 7 by repeatedly adding 7. The board writes 1 ≡ 8 ≡ 15 (mod 7). Since 8 is not divisible by 5 but 15 is, the number 15 is factored as 5⋅35\cdot 3. From 5⋅35\cdot 3 ≡ 1 (mod 7), the inverse is read off as 3, and the result is written as 5−15^{-1} ≡ 3 (mod 7).

A second example asks for the inverse of 5 modulo 11. The same search strategy is used, but now the modulus is 11, so the chain becomes longer: 1 ≡ 12 ≡ 23 ≡ 34 ≡ 45 (mod 11). The first multiple of 5 encountered in this sequence is 45, which factors as 5⋅95\cdot 9. Therefore 5⋅95\cdot 9 ≡ 1 (mod 11), and the inverse is 9, written as 5−15^{-1} ≡ 9 (mod 11).

The final example finds the inverse of 6 modulo 17. Here only one addition of the modulus is needed: 1 ≡ 18 (mod 17). Since 18=6⋅318 = 6\cdot 3, we have 6⋅36\cdot 3 ≡ 1 (mod 17), so the inverse of 6 modulo 17 is 3, written as 6−16^{-1} ≡ 3 (mod 17). Across all three examples, the visual procedure consistently demonstrates the same principle: locate a representative congruent to 1 modulo m that is divisible by the target integer, then factor it to identify the inverse.

The clip opens on a digital whiteboard after a previous example, then switches to the new typed prompt: Find the inverse of 18 mod 19.

The instructor begins with the most direct method. Since an inverse of 18 modulo 19 is a number b with 18b ≡ 1 (mod 19), they generate the congruence class of 1 by repeatedly adding 19: 1 ≡ 20 ≡ 39 ≡ 58 ≡ 77 ≡ ... 324 (mod 19).

When the chain reaches 324, the board factors it as 18⋅1818\cdot 18. That means 18⋅1818\cdot 18 ≡ 1 (mod 19), so the inverse of 18 modulo 19 is 18 itself.

The speaker notes that this brute-force procedure always works, but it can waste time because you may have to add the modulus many times before hitting a multiple of the number whose inverse you want.

A shortcut is then introduced: choose a more convenient representative in the same residue class. Since 18−19=−118 - 19 = -1, we have 18 ≡ -1 (mod 19).

Now the inverse question becomes trivial. Because (-1)(-1)=1, the residue -1 is its own inverse. Therefore 18, being congruent to -1 modulo 19, is also its own inverse: 18−118^{-1} ≡ 18 (mod 19).

The same idea is applied to a second example: find the inverse of 17 mod 19. First rewrite 17 as a negative residue: 17 ≡ -2 (mod 19), since -2 + 19=1719 = 17.

The board then reuses the earlier fact that 20 ≡ 1 (mod 19). Because 20=(−2)(−10)20 = (-2)(-10), it follows that (-2)^{-1} ≡ -10 (mod 19).

Finally, the negative answer is converted back to a positive representative by adding 19: -10 + 19=919 = 9. Hence 17−117^{-1} ≡ 9 (mod 19).

The completed board leaves both results visible: 18−118^{-1} ≡ 18 (mod 19) and 17−117^{-1} ≡ 9 (mod 19), illustrating that negative residues often make modular inverse calculations much faster.

The clip opens on a finished modulo-19 example. The printed prompt asks for the inverse of 18 mod 19. Beneath it, the board records 18≡−1(mod19)18 \equiv -1 \pmod{19} and then 18−1≡1(mod19)18^{-1} \equiv 1 \pmod{19}. The mathematical point is that reducing 18 to the simpler residue -1 makes the inverse immediate, since -1 is its own inverse modulo 19.

The same board also contains a second completed task, "Find inverse of 17 mod 19." The written chain is 17≡−2(mod19)17 \equiv -2 \pmod{19}, then (-2)^{-1} ≡−10(mod19)\equiv -10 \pmod{19}, then 17−1≡9(mod19)17^{-1} \equiv 9 \pmod{19}. This shows the method of replacing 17 by a smaller negative residue, inverting that residue, and finally rewriting the answer as a positive residue modulo 19.

While these results are already on screen, the speaker comments on the slower brute-force route: if one kept extending the list of congruences, one would eventually arrive at 17×9=15317 \times 9 = 153. The audio therefore contrasts a long search with the shorter structural method already written on the board.

The display changes to a new page with the printed question "Find the inverse of 5 mod 10." This transition signals a shift from examples where inverses exist to a case where they do not.

The presenter begins a direct search by writing 1≡11≡21≡31≡⋯1 \equiv 11 \equiv 21 \equiv 31 \equiv \cdots modulo 10. The idea is to list numbers congruent to 1 modulo 10 by repeatedly adding 10. The audio explains that this process could continue indefinitely without ever producing a multiple of 5, so the desired inverse cannot be found this way.

The board then records the decisive obstruction: (5,10)=5 NOT 1. Here (5,10) denotes the greatest common divisor of 5 and 10. Because that gcd is 5 rather than 1, the presenter writes the conclusion No inverse underneath.

The spoken rule is generalized beyond this single example: if the greatest common divisor of the two values is greater than 1, then there is no inverse. The clip closes with brief encouragement to practice, with no further mathematics added.

Knowledge cards

01

Multiplicative inverse modulo m

Two integers a and b are multiplicative inverses modulo m when their product is congruent to 1 modulo m. This is the basic object studied in the clip.

ab≡1(modm)ab \equiv 1 \pmod{m}
02

Existence theorem for modular inverses

An integer a has an inverse modulo m if and only if gcd(a,m)=1. If the gcd is greater than 1, no inverse exists.

a has an inverse modulo m  ⟺  (a,m)=1a \text{ has an inverse modulo } m \iff (a,m)=1
03

Search method for inverses

To find an inverse of a modulo m by hand, start from 1 and keep adding m until reaching a multiple of a. Factor that multiple as a⋅ba\cdot b; then b is the inverse modulo m.

1≡1+m≡1+2m≡⋯(modm)1 \equiv 1+m \equiv 1+2m \equiv \cdots \pmod{m}
04

Example: inverse of 5 mod 7

Using the search method, 1 ≡ 8 ≡ 15 (mod 7), and 15=5⋅315 = 5\cdot 3, so the inverse of 5 modulo 7 is 3.

5−1≡3(mod7)5^{-1} \equiv 3 \pmod{7}
05

Example: inverse of 5 mod 11

Using the search method, 1 ≡ 12 ≡ 23 ≡ 34 ≡ 45 (mod 11), and 45=5⋅945 = 5\cdot 9, so the inverse of 5 modulo 11 is 9.

5−1≡9(mod11)5^{-1} \equiv 9 \pmod{11}
06

Example: inverse of 6 mod 17

Using the search method, 1 ≡ 18 (mod 17), and 18=6⋅318 = 6\cdot 3, so the inverse of 6 modulo 17 is 3.

6−1≡3(mod17)6^{-1} \equiv 3 \pmod{17}
07

Modular inverse definition

For integers a and modulus m, the modular inverse of a is a residue b such that ab ≡ 1 (mod m). In this clip the modulus is 19, so finding the inverse of 18 means finding b with 18b ≡ 1 (mod 19).

aa−1≡1(modm)aa^{-1} \equiv 1 \pmod m
08

Brute-force method by adding the modulus

One reliable but slow strategy is to list numbers congruent to 1 modulo m by repeatedly adding m, then stop when one is divisible by a. The video shows 1 ≡ 20 ≡ 39 ≡ 58 ≡ 77 ≡ ... ≡ 324 (mod 19), and 324=18⋅18324 = 18\cdot 18, so 18−118^{-1} ≡ 18 (mod 19).

1+km≡1(modm)1+km \equiv 1 \pmod m
09

Negative-residue shortcut

Instead of staying with large positive representatives, replace a by a congruent negative residue. Since 18−19=−118-19=-1, the problem of inverting 18 modulo 19 becomes the problem of inverting -1, which is immediate.

a≡a−m(modm)a \equiv a-m \pmod m
10

Why -1 is its own inverse

Because (-1)(-1)=1, the residue -1 is always its own multiplicative inverse. Applying this modulo 19 gives 18 ≡ -1 (mod 19), hence 18−118^{-1} ≡ 18 (mod 19).

(−1)2≡1(modm)(-1)^2 \equiv 1 \pmod m
11

Inverse of 17 modulo 19 via -2

Rewrite 17 as -2 modulo 19. From the earlier line 20 ≡ 1 (mod 19) and the factorization 20=(−2)(−10)20 = (-2)(-10), we get (-2)^{-1} ≡ -10 (mod 19). Adding 19 converts -10 to the positive residue 9, so 17−117^{-1} ≡ 9 (mod 19).

17≡−2(mod19),  (−2)−1≡−10(mod19),  17−1≡9(mod19)17 \equiv -2 \pmod{19},\; (-2)^{-1} \equiv -10 \pmod{19},\; 17^{-1} \equiv 9 \pmod{19}
12

Convert negative answers to positive residues

A negative residue is a valid intermediate answer, but final answers are often written as positive representatives. The clip demonstrates this by changing -10 into 9 using -10 + 19=919 = 9.

x≡x+m(modm)x \equiv x+m \pmod m
13

Modular inverse notation used in the lesson

Throughout the clip, "the inverse of a mod m" is written as a−1≡x(modm)a^{-1} \equiv x \pmod m. The examples make clear that the topic is multiplicative inversion in modular arithmetic: one seeks a residue x modulo m that serves as the inverse of a modulo m. The clip does not separately spell out the defining equation in words, but the notation and worked cases consistently use this meaning.

a−1≡x(modm)a^{-1} \equiv x \pmod m
14

Inverse of 18 modulo 19

The board rewrites 18 as -1 modulo 19 and then immediately records the inverse as 1 modulo 19. This is a compact example of simplifying the representative before inverting.

18≡−1(mod19),18−1≡1(mod19)18 \equiv -1 \pmod{19},\quad 18^{-1} \equiv 1 \pmod{19}
15

Inverse of 17 modulo 19

The presenter reduces 17 to -2 modulo 19, uses the displayed fact (-2)^{-1} ≡−10(mod19)\equiv -10 \pmod{19}, and then rewrites -10 as 9 modulo 19. The audio adds that a longer search would also reach the product 17×9=15317 \times 9 = 153.

17≡−2(mod19),(−2)−1≡−10(mod19),17−1≡9(mod19)17 \equiv -2 \pmod{19},\quad (-2)^{-1} \equiv -10 \pmod{19},\quad 17^{-1} \equiv 9 \pmod{19}
16

Failed direct search for the inverse of 5 modulo 10

To test whether 5 has an inverse modulo 10, the board lists numbers congruent to 1 modulo 10: 1, 11, 21, 31, and so on. The spoken explanation is that repeatedly adding 10 never produces a multiple of 5, so this route cannot yield an inverse.

1≡11≡21≡31≡⋯(mod10)1 \equiv 11 \equiv 21 \equiv 31 \equiv \cdots \pmod{10}
17

GCD criterion shown for nonexistence of an inverse

The decisive observation in the second example is that (5,10)=5, which is not 1. From this the board concludes No inverse. The speaker then states the broader rule that if the greatest common divisor of the two values is greater than 1, then there is no inverse.

(5,10)=5≠1⇒No inverse(5,10)=5 \ne 1 \Rightarrow \text{No inverse}

Detailed learning notes

Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.

Symbols · 18

a

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The definition line uses the integer symbol a in “integers a and b are multiplicative inverses modulo m if ab ≡ 1 (mod m).”

  2. Formula
    Observation

    The theorem line uses the same symbol a in “The integer a has an inverse modulo m if and only if (a,m)=1.”

Symbol

a

Meaning

An integer whose multiplicative inverse modulo m is being discussed; in the examples it is instantiated as 5 or 6.

Domain

Integers.

b

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The definition line uses b in “integers a and b are multiplicative inverses modulo m if ab ≡ 1 (mod m).”

Symbol

b

Meaning

An integer that serves as a multiplicative inverse of a modulo m when ab ≡ 1 (mod m).

Domain

Integers.

m

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The definition line contains “modulo m” and “(mod m)”.

  2. Formula
    Observation

    The theorem line contains “inverse modulo m” and “(a,m)=1”.

Symbol

m

Meaning

The modulus for the congruence relation and for the existence condition of the inverse.

Domain

Integers used as a modulus; the video does not explicitly state extra restrictions on m.

(a,m)

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The theorem line shows “(a,m)=1”.

  2. Audio
    Observation

    The speaker reads this as “the greatest common divisor of a and m is equal to 1.”

Symbol

(a,m)

Meaning

The greatest common divisor of a and m.

Domain

Defined for the integer pair a and m appearing in the theorem.

≡

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The definition uses “ab ≡ 1 (mod m)”.

  2. Formula
    Observation

    The worked examples use chains such as “1 ≡ 8 ≡ 15 (mod 7)” and “1 ≡ 12 ≡ 23 ≡ 34 ≡ 45 (mod 11)”.

Symbol

≡

Meaning

Congruence modulo the stated modulus.

Domain

Used between integers with a specified modulus in parentheses.

x−1x^{-1}

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The first example writes “5−15^{-1} ≡ 3 (mod 7)”.

  2. Formula
    Observation

    The second example writes “5−15^{-1} ≡ 9 (mod 11)”.

  3. Formula
    Observation

    The third example writes “6−16^{-1} ≡ 3 (mod 17)”.

Symbol

x−1x^{-1}

Meaning

Notation for the multiplicative inverse of x modulo the given modulus.

Domain

Used for integers x in the examples, namely 5 and 6.

19

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The modulus 19 appears in the problem title and in congruence statements such as 18 ≡ -1 (mod 19) and 17 ≡ -2 (mod 19).

Symbol

19

Meaning

The modulus used throughout the main worked examples in this clip.

Domain

Positive integer modulus.

≡

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The handwritten symbol '≡' is used repeatedly to state congruences modulo 19.

Symbol

≡

Meaning

Congruence relation; a ≡ b (mod m) means a and b have the same remainder modulo m.

Domain

Binary relation on integers.

a−1a^{-1}

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Notation such as 18−118^{-1}, (-2)^{-1}, and 17−117^{-1} is written on the board.

Symbol

a−1a^{-1}

Meaning

Multiplicative inverse of a modulo the stated modulus.

Domain

Defined when gcd(a,m)=1; here the relevant residues are invertible modulo 19.

18

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The number 18 is the residue whose inverse modulo 19 is being found.

Symbol

18

Meaning

Residue class modulo 19; also identified as -1 modulo 19.

Domain

Integer representative of a residue class mod 19.

17

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The second example asks for the inverse of 17 mod 19.

Symbol

17

Meaning

Residue class modulo 19; also identified as -2 modulo 19.

Domain

Integer representative of a residue class mod 19.

-1

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board writes 18 ≡ -1 (mod 19).

Symbol

-1

Meaning

Negative residue equivalent to 18 modulo 19.

Domain

Residue class modulo 19.

Knowledge points · 11

Definition of multiplicative inverses modulo m

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    On-screen text: “Defn: We say that integers a and b are multiplicative inverses modulo m if ab ≡ 1 (mod m).”

  2. Audio
    Observation

    The speaker says, “So we're finding multiplicative inverses, and that's two numbers that multiply together to get you one mod m.”

Definition
Explanation

The video defines a pair of integers a and b as multiplicative inverses modulo m exactly when their product is congruent to 1 modulo m. The speaker paraphrases this as “two numbers that multiply together to get you one mod m.”

Formula
ab≡1(modm)ab \equiv 1 \pmod{m}
Conditions
  1. a and b are integers.

  2. The congruence is taken modulo m.

Existence criterion for a modular inverse

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    On-screen text: “Theorem: The integer a has an inverse modulo m if and only if (a,m)=1.”

  2. Audio
    Observation

    The speaker says, “a has an inverse mod m if and only if the greatest common divisor of a and m is equal to 1.”

  3. Audio
    Observation

    The speaker adds, “If the greatest common divisor between a and m is greater than 1, then there is no inverse mod m.”

Method
Explanation

The clip states a necessary-and-sufficient condition for the existence of a modular inverse: an integer a has an inverse modulo m precisely when gcd(a,m)=1. The speaker also states the contrapositive in words: if gcd(a,m)>1, then no inverse modulo m exists.

Formula
a has an inverse modulo m  ⟺  (a,m)=1a \text{ has an inverse modulo } m \iff (a,m)=1
Conditions
  1. a is an integer.

  2. m is the modulus.

  3. (a,m) denotes gcd(a,m).

Prerequisites
  1. Definition of multiplicative inverses modulo m

Search method for modular inverses by adding the modulus

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    In the first example the speaker says, “So we do as we've done in the past. Mod 7, 1 is congruent to 8. That's not a multiple of 5. So we add 7 again and we get 15.”

  2. Audio
    Observation

    In the second example the speaker says, “Always start with 1 and add the mod on.”

  3. Formula
    Observation

    The written chains are “1 ≡ 8 ≡ 15 (mod 7)”, “1 ≡ 12 ≡ 23 ≡ 34 ≡ 45 (mod 11)”, and “1 ≡ 18 (mod 17)”.

Method
Explanation

The worked examples use a trial method: start from 1 and repeatedly add the modulus m until reaching a number divisible by the target integer a. Once such a representative is found, factor it as a times another integer; that other integer is the inverse modulo m.

Formula
1≡1+m≡1+2m≡⋯(modm)1 \equiv 1+m \equiv 1+2m \equiv \cdots \pmod{m}
Conditions
  1. Used to find an inverse of a modulo m.

  2. One looks among integers congruent to 1 modulo m for a multiple of a.

Prerequisites
  1. Definition of multiplicative inverses modulo m

Modular multiplicative inverse

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker repeatedly says 'find the inverse of 18 mod 19' and later 'find the inverse of 17 mod 19'.

  2. Formula
    Observation

    The board writes conclusions such as 18−118^{-1} ≡ 18 and 17−117^{-1} ≡ 9 (mod 19).

Definition
Explanation

For a residue a modulo m, its inverse a−1a^{-1} is the residue b such that ab ≡ 1 (mod m). The clip uses this notion directly when searching for numbers whose product with 18 or 17 is congruent to 1 modulo 19.

Formula
aa−1≡1(modm)aa^{-1} \equiv 1 \pmod m
Conditions
  1. The inverse exists when a is relatively prime to m.

  2. In this clip the modulus is m=19m=19.

Prerequisites
  1. ≡
  2. a−1a^{-1}

Brute-force search by adding the modulus

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says they can start with 1 ≡ 20, then keep adding 19 until reaching a multiple of 18.

  2. Formula
    Observation

    The board writes 1 ≡ 20 ≡ 39 ≡ 58 ≡ 77 ≡ ... ≡ 324, then factors 324 as 18⋅1818\cdot 18.

Method
Explanation

One way to find an inverse modulo m is to list numbers congruent to 1 modulo m by repeatedly adding m, then stop when one of those numbers is divisible by a. In the example, the sequence 20, 39, 58, 77, ..., 324 consists of numbers congruent to 1 modulo 19, and 324=18⋅18324 = 18\cdot 18 gives the inverse.

Formula
1+km≡1(modm),stop when 1+km=ab1 + km \equiv 1 \pmod m,\quad \text{stop when } 1+km = a b
Conditions
  1. Works whenever a has an inverse modulo m.

  2. May be slow if many terms must be checked.

Prerequisites
  1. Modular multiplicative inverse
  2. ≡

Use negative residues to simplify inverse problems

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker introduces 'a little bit of a shortcut' by noticing 18 is the same thing as negative 1 mod 19, and later 17 is the same thing as negative 2 mod 19.

  2. Formula
    Observation

    The board writes 18 ≡ -1 (mod 19) and 17 ≡ -2 (mod 19).

Method
Explanation

Instead of working only with positive representatives, replace a by an equivalent smaller-magnitude residue modulo m. This often makes the inverse easier to recognize. Here 18 is replaced by -1 and 17 by -2, turning the problem into finding inverses of simpler residues.

Formula
a≡a−m(modm)a \equiv a-m \pmod m
Conditions
  1. The replacement must be congruent modulo the same m.

  2. The resulting residue should still be invertible modulo m.

Prerequisites
  1. Modular multiplicative inverse
  2. ≡

The residue -1 is its own inverse

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker asks what the inverse of negative 1 is and answers that multiplying negative 1 by negative 1 gives positive 1.

  2. Formula
    Observation

    The board shows 18 ≡ -1 (mod 19) and concludes 18−118^{-1} ≡ 18 (mod 19).

Formula
Explanation

Because (-1)(-1)=1, the residue -1 modulo any m is always its own multiplicative inverse. Since 18 ≡ -1 (mod 19), it follows immediately that 18−118^{-1} ≡ 18 (mod 19).

Formula
(−1)2≡1(modm)⇒(−1)−1≡−1(modm)(-1)^2 \equiv 1 \pmod m \Rightarrow (-1)^{-1} \equiv -1 \pmod m
Conditions
  1. Valid for any modulus m.

Prerequisites
  1. Modular multiplicative inverse
  2. Use negative residues to simplify inverse problems

Convert a negative inverse back to a positive residue

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says they will add 19 to negative 10 because they want positive numbers.

  2. Formula
    Observation

    The board converts (-2)^{-1} ≡ -10 (mod 19) into 17−117^{-1} ≡ 9 (mod 19).

Method
Explanation

After finding an inverse in negative form, add the modulus once or more to obtain the standard positive representative. Here -10 + 19=919 = 9, so the inverse of 17 modulo 19 is reported as 9.

Formula
x≡x+m(modm)x \equiv x+m \pmod m
Conditions
  1. Used when a negative residue answer is acceptable algebraically but a positive representative is preferred.

Prerequisites
  1. ≡
  2. Use negative residues to simplify inverse problems

Worked example: inverses of 18 and 17 modulo 19

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Printed title: "Find the inverse of 18 mod 19."

  2. Formula
    Observation

    Written lines include 18≡−1(mod19)18 \equiv -1 \pmod{19} and 18−1≡1(mod19)18^{-1} \equiv 1 \pmod{19}.

  3. Formula
    Observation

    Written lines include 17≡−2(mod19)17 \equiv -2 \pmod{19}, (-2)^{-1} ≡−10(mod19)\equiv -10 \pmod{19}, and 17−1≡9(mod19)17^{-1} \equiv 9 \pmod{19}.

Method
Explanation

This segment shows two completed modular-inverse computations modulo 19. For 18, the presenter rewrites 18 as -1 modulo 19 and concludes that its inverse is 1 modulo 19. For 17, the presenter rewrites 17 as -2 modulo 19, uses (-2)^{-1} ≡−10(mod19)\equiv -10 \pmod{19}, and then records 17−1≡9(mod19)17^{-1} \equiv 9 \pmod{19}. The audio adds that the same result for 17 could also be reached by continuing a longer list of congruences until reaching 17×9=15317 \times 9 = 153.

Formula
18≡−1(mod19),18−1≡1(mod19);17≡−2(mod19),(−2)−1≡−10(mod19),17−1≡9(mod19)18 \equiv -1 \pmod{19},\quad 18^{-1} \equiv 1 \pmod{19};\qquad 17 \equiv -2 \pmod{19},\quad (-2)^{-1} \equiv -10 \pmod{19},\quad 17^{-1} \equiv 9 \pmod{19}
Conditions
  1. Modulus is 19.

  2. The displayed method assumes the relevant inverses already exist modulo 19.

Prerequisites
  1. Meaning of a modular inverse in this lesson

Meaning of a modular inverse in this lesson

Clear evidence
Derived from the video
Evidence
  1. Formula
    Observation

    The board repeatedly uses notation a−1≡b(modm)a^{-1} \equiv b \pmod m in examples such as 18−1≡1(mod19)18^{-1} \equiv 1 \pmod{19} and 17−1≡9(mod19)17^{-1} \equiv 9 \pmod{19}.

  2. Audio
    Observation

    The speaker frames both examples as "find the inverse" problems modulo a given number.

Definition
Explanation

In this video, "the inverse of a mod m" means a residue class x modulo m that serves as a multiplicative inverse of a modulo m. The worked examples use the notation a−1≡x(modm)a^{-1} \equiv x \pmod m. The clip does not separately state the defining equation in words, but the notation and examples make clear that the topic is multiplicative inversion in modular arithmetic.

Formula
a−1≡x(modm)a^{-1} \equiv x \pmod m
Conditions
  1. Applies to integers a and positive modulus m.

  2. Existence is not automatic; the later example shows failure when gcd(a,m) ≠1\ne 1.

Criterion for nonexistence of a modular inverse

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board writes (5,10)=5 NOT 1 and then No inverse.

  2. Audio
    Observation

    The speaker says, "the greatest common divisor of 5 and 10 is 5, which is not 1. So the answer is that there is no inverse."

  3. Audio
    Observation

    The speaker generalizes: "if there's a greatest common divisor greater than 1 between those two values, then there is no inverse."

Method
Explanation

The second example teaches a quick existence test before trying to compute an inverse. For 5 mod 10, the presenter computes (5,10)=5, notes that this is not 1, and concludes that 5 has no inverse modulo 10. The spoken rule is generalized to any pair whose greatest common divisor is greater than 1.

Formula
(5,10)=5≠1⇒No inverse(5,10)=5 \ne 1 \Rightarrow \text{No inverse}
Conditions
  1. Used for deciding whether a modular inverse exists.

  2. In this clip the rule is stated for the case gcd > 1 implying no inverse.

Prerequisites
  1. Meaning of a modular inverse in this lesson
Claims and conditions · 8

Integer a has an inverse modulo m iff gcd(a,m)=1

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The theorem statement is visible on screen: “The integer a has an inverse modulo m if and only if (a,m)=1.”

  2. Audio
    Observation

    The speaker reads the theorem aloud and explains the case gcd(a,m)>1 means no inverse exists.

Theorem
Statement

For integers a and m, a has a multiplicative inverse modulo m if and only if (a,m)=1.

Hypotheses
  1. a is an integer.

  2. m is the modulus.

Quantifiers

For the integers a and m appearing in the statement.

If gcd(a,m)>1 then no inverse modulo m exists

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says, “If the greatest common divisor between a and m is greater than 1, then there is no inverse mod m.”

Proposition
Statement

If the greatest common divisor of a and m is greater than 1, then a has no inverse modulo m.

Hypotheses
  1. a and m are integers.

  2. (a,m)>1.

Quantifiers

For integers a and m satisfying the gcd condition.

Inverse of 18 modulo 19

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board writes 18−118^{-1} ≡ 18 after showing 324=18⋅18324 = 18\cdot 18, and again after using 18 ≡ -1 (mod 19).

  2. Audio
    Observation

    The speaker states, 'So 18's inverse is 18' and later '18 is its own inverse.'

Proposition
Statement

18−118^{-1} ≡ 18 (mod 19).

Hypotheses
  1. Work modulo 19.

  2. Use the fact that 18 ≡ -1 (mod 19), or equivalently that 18⋅18=32418\cdot 18 = 324 ≡ 1 (mod 19).

Quantifiers

Specific numerical statement for modulus 19.

Inverse of 17 modulo 19

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board writes 17 ≡ -2 (mod 19), then (-2)^{-1} ≡ -10 (mod 19), then 17−117^{-1} ≡ 9 (mod 19).

  2. Audio
    Observation

    The speaker concludes that the inverse of 17 is congruent to 9 mod 19.

Proposition
Statement

17−117^{-1} ≡ 9 (mod 19).

Hypotheses
  1. Work modulo 19.

  2. Use 17 ≡ -2 (mod 19) and 20=(−2)(−10)20 = (-2)(-10), so (-2)^{-1} ≡ -10 (mod 19).

  3. Convert -10 to the positive representative 9 modulo 19.

Quantifiers

Specific numerical statement for modulus 19.

Negative one is its own inverse

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker asks what multiplied by negative 1 gives positive 1 and answers negative 1.

  2. Formula
    Observation

    The reasoning is written through 18 ≡ -1 (mod 19) and the conclusion 18−118^{-1} ≡ 18 (mod 19).

Proposition
Statement

(-1)^{-1} ≡ -1 (mod m), because (-1)(-1)=1.

Hypotheses
  1. Any modulus m is allowed for this general fact.

  2. In the clip it is applied with m=19m=19.

Quantifiers

General proposition applied to modulus 19 in the example.

If gcd(a,m)>1, then a has no inverse modulo m

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    "So if there's a greatest common divisor greater than 1 between those two values, then there is no inverse."

  2. Formula
    Observation

    Board text: (5,10)=5 NOT 1, followed by No inverse.

Uncertainties
  1. The clip states only the direction gcd>1 implies no inverse; it does not explicitly state the converse condition gcd=1 implies existence.

Proposition
Statement

For the numbers under consideration, if the greatest common divisor of a and m is greater than 1, then a has no modular inverse modulo m.

Hypotheses
  1. a and m are integers with modulus m.

  2. (a,m)>1.

Quantifiers

The spoken wording is general over "those two values," but the clip only proves or illustrates the claim in the specific example a=5a=5, m=10m=10.

The inverse of 17 modulo 19 is 9

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Board text includes 17−1≡9(mod19)17^{-1} \equiv 9 \pmod{19}.

  2. Audio
    Observation

    The speaker says that eventually one would get to 17 times 9, which would be 153.

Proposition
Statement

17−1≡9(mod19)17^{-1} \equiv 9 \pmod{19}.

Hypotheses
  1. Working modulo 19.

  2. Using the intermediate facts 17≡−2(mod19)17 \equiv -2 \pmod{19} and (-2)^{-1} ≡−10(mod19)\equiv -10 \pmod{19} shown on the board.

Quantifiers

Specific numerical claim.

The inverse of 18 modulo 19 is 1

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Board text includes 18≡−1(mod19)18 \equiv -1 \pmod{19} and 18−1≡1(mod19)18^{-1} \equiv 1 \pmod{19}.

Proposition
Statement

18−1≡1(mod19)18^{-1} \equiv 1 \pmod{19}.

Hypotheses
  1. Working modulo 19.

  2. Using 18≡−1(mod19)18 \equiv -1 \pmod{19}.

Quantifiers

Specific numerical claim.

Derivations and proofs · 9

Derivation of the inverse of 5 modulo 7

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board shows the chain “1 ≡ 8 ≡ 15 (mod 7)” with 15 factored below into 5 and 3.

  2. Audio
    Observation

    The speaker says, “Mod 7, 1 is congruent to 8. That's not a multiple of 5. So we add 7 again and we get 15. 15 is 5 times 3 mod 7.”

  3. Formula
    Observation

    The final written conclusion is “5−15^{-1} ≡ 3 (mod 7)”.

Numerical verification
Steps
  1. Expression
    1≡8(mod7)1 \equiv 8 \pmod{7}
    Explanation

    Start from 1 and add the modulus 7 once to obtain another representative congruent to 1 modulo 7.

    Justification

    Adding the modulus preserves congruence modulo 7.

    Shown in the video
  2. Expression
    8≢0(mod5)8 \not\equiv 0 \pmod{5}
    Explanation

    Check whether 8 is a multiple of 5; it is not, so continue searching.

    Justification

    The method requires finding a representative congruent to 1 modulo 7 that is divisible by 5.

    Derived from the video
  3. Expression
    1≡8≡15(mod7)1 \equiv 8 \equiv 15 \pmod{7}
    Explanation

    Add 7 again to reach 15, which is still congruent to 1 modulo 7.

    Justification

    Repeated addition of the modulus stays within the same congruence class.

    Shown in the video
  4. Expression
    15=5⋅315 = 5 \cdot 3
    Explanation

    Factor 15 as 5 times 3.

    Justification

    Ordinary integer factorization.

    Shown in the video
  5. Expression
    5−1≡3(mod7)5^{-1} \equiv 3 \pmod{7}
    Explanation

    Since 5⋅35\cdot 3 ≡ 15 ≡ 1 (mod 7), the inverse of 5 modulo 7 is 3.

    Justification

    By the definition of multiplicative inverse modulo m.

    Shown in the video
Conclusion

The inverse of 5 modulo 7 is 3, written as 5−15^{-1} ≡ 3 (mod 7).

Derivation of the inverse of 5 modulo 11

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board shows “1 ≡ 12 ≡ 23 ≡ 34 ≡ 45 (mod 11)” with 45 factored into 5 and 9.

  2. Audio
    Observation

    The speaker says, “So hopefully you said 1 was congruent to 12 mod 11. Always start with 1 and add the mod on. Congruent to 23, adding 11, 34, and eventually we get to 45. And that's 5 times 9 mod 11.”

  3. Formula
    Observation

    The final written conclusion is “5−15^{-1} ≡ 9 (mod 11)”.

Numerical verification
Steps
  1. Expression
    1≡12(mod11)1 \equiv 12 \pmod{11}
    Explanation

    Begin with 1 and add the modulus 11 to get 12.

    Justification

    Adding the modulus preserves congruence modulo 11.

    Shown in the video
  2. Expression
    12≡23≡34≡45(mod11)12 \equiv 23 \equiv 34 \equiv 45 \pmod{11}
    Explanation

    Continue adding 11 until reaching a number divisible by 5.

    Justification

    All these numbers remain congruent to 1 modulo 11.

    Shown in the video
  3. Expression
    45=5⋅945 = 5 \cdot 9
    Explanation

    Factor 45 as 5 times 9.

    Justification

    Ordinary integer factorization.

    Shown in the video
  4. Expression
    5−1≡9(mod11)5^{-1} \equiv 9 \pmod{11}
    Explanation

    Because 5⋅9=455\cdot 9 = 45 ≡ 1 (mod 11), the inverse of 5 modulo 11 is 9.

    Justification

    By the definition of multiplicative inverse modulo m.

    Shown in the video
Conclusion

The inverse of 5 modulo 11 is 9, written as 5−15^{-1} ≡ 9 (mod 11).

Derivation of the inverse of 6 modulo 17

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board shows “1 ≡ 18 (mod 17)” with 18 factored into 6 and 3.

  2. Audio
    Observation

    The speaker says, “So you've got 1 congruent to 18 because we're adding 17 this time, and that's 6 times 3 mod 17.”

  3. Formula
    Observation

    The final written conclusion is “6−16^{-1} ≡ 3 (mod 17)”.

Numerical verification
Steps
  1. Expression
    1≡18(mod17)1 \equiv 18 \pmod{17}
    Explanation

    Start from 1 and add the modulus 17 to obtain 18.

    Justification

    Adding the modulus preserves congruence modulo 17.

    Shown in the video
  2. Expression
    18=6⋅318 = 6 \cdot 3
    Explanation

    Factor 18 as 6 times 3.

    Justification

    Ordinary integer factorization.

    Shown in the video
  3. Expression
    6−1≡3(mod17)6^{-1} \equiv 3 \pmod{17}
    Explanation

    Since 6⋅3=186\cdot 3 = 18 ≡ 1 (mod 17), the inverse of 6 modulo 17 is 3.

    Justification

    By the definition of multiplicative inverse modulo m.

    Shown in the video
Conclusion

The inverse of 6 modulo 17 is 3, written as 6−16^{-1} ≡ 3 (mod 17).

Deriving 18−118^{-1} ≡ 18 by brute force

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board lists 1 ≡ 20 ≡ 39 ≡ 58 ≡ 77 ≡ ... ≡ 324 and factors 324 as 18⋅1818\cdot 18.

  2. Audio
    Observation

    The speaker narrates adding 19 repeatedly until reaching 324, which is 18 times 18.

Uncertainties
  1. The intermediate omitted terms between 77 and 324 are not all written explicitly; only the ellipsis is shown.

Proof
Steps
  1. Expression
    1≡20≡39≡58≡77≡⋯≡324(mod19)1 \equiv 20 \equiv 39 \equiv 58 \equiv 77 \equiv \cdots \equiv 324 \pmod{19}
    Explanation

    Start from 1 and keep adding the modulus 19 to generate numbers congruent to 1 modulo 19.

    Justification

    If x ≡ 1 (mod 19), then x+19x+19 ≡ 1 (mod 19).

    Shown in the video
  2. Expression
    324=18⋅18324 = 18 \cdot 18
    Explanation

    Identify the first listed number that is a multiple of 18.

    Justification

    Direct factorization shown on the board.

    Shown in the video
  3. Expression
    18−1≡18(mod19)18^{-1} \equiv 18 \pmod{19}
    Explanation

    Since 18⋅1818\cdot 18 ≡ 1 (mod 19), the inverse of 18 modulo 19 is 18.

    Justification

    Definition of modular multiplicative inverse.

    Shown in the video
Conclusion

The brute-force search proves that 18−118^{-1} ≡ 18 (mod 19).

Shortcut derivation of 18−118^{-1} using -1

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board writes 18 ≡ -1 (mod 19) and then 18−118^{-1} ≡ 18 (mod 19).

  2. Audio
    Observation

    The speaker explains that instead of adding 19, we can subtract 19, and that the inverse of -1 is -1.

Proof
Steps
  1. Expression
    18≡−1(mod19)18 \equiv -1 \pmod{19}
    Explanation

    Replace 18 by the congruent negative residue -1.

    Justification

    Subtracting the modulus preserves congruence: 18−19=−118-19=-1.

    Shown in the video
  2. Expression
    (−1)(−1)=1(-1)(-1)=1
    Explanation

    Recognize that -1 multiplied by itself gives 1.

    Justification

    Basic integer multiplication.

    Shown in the video
  3. Expression
    18−1≡18(mod19)18^{-1} \equiv 18 \pmod{19}
    Explanation

    Therefore the inverse of 18 is itself modulo 19.

    Justification

    Because 18 ≡ -1 and (-1)^{-1} ≡ -1.

    Shown in the video
Conclusion

Using the negative residue -1 gives a much faster proof that 18 is its own inverse modulo 19.

Shortcut derivation of 17−117^{-1} using -2

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board writes 17 ≡ -2 (mod 19), notes 20=(−2)(−10)20 = (-2)(-10), writes (-2)^{-1} ≡ -10 (mod 19), and finally 17−117^{-1} ≡ 9 (mod 19).

  2. Audio
    Observation

    The speaker says 17 is the same thing as negative 2 mod 19, looks back at 20 = negative 2 times negative 10, and then adds 19 to negative 10 to get a positive number.

Proof
Steps
  1. Expression
    17≡−2(mod19)17 \equiv -2 \pmod{19}
    Explanation

    Rewrite 17 as the negative residue -2.

    Justification

    Adding 19 to -2 gives 17, so they are congruent modulo 19.

    Shown in the video
  2. Expression
    20=(−2)(−10)20 = (-2)(-10)
    Explanation

    Use the earlier line 1 ≡ 20 (mod 19) and factor 20 as (-2)(-10).

    Justification

    Direct arithmetic identity shown on the board.

    Shown in the video
  3. Expression
    (−2)−1≡−10(mod19)(-2)^{-1} \equiv -10 \pmod{19}
    Explanation

    Since (-2)(-10)=20 ≡ 1 (mod 19), -10 is the inverse of -2 modulo 19.

    Justification

    Definition of modular inverse together with 20 ≡ 1 (mod 19).

    Shown in the video
  4. Expression
    −10+19=9-10 + 19 = 9
    Explanation

    Convert the negative representative -10 to the positive residue 9.

    Justification

    Adding the modulus does not change the congruence class.

    Shown in the video
  5. Expression
    17−1≡9(mod19)17^{-1} \equiv 9 \pmod{19}
    Explanation

    Therefore the inverse of 17 modulo 19 is 9.

    Justification

    Substitute 17 ≡ -2 and the converted inverse.

    Shown in the video
Conclusion

The shortcut method yields 17−117^{-1} ≡ 9 (mod 19).

Derivation of 18−118^{-1} modulo 19

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board shows 18≡−1(mod19)18 \equiv -1 \pmod{19} and directly below it 18−1≡1(mod19)18^{-1} \equiv 1 \pmod{19}.

Proof
Steps
  1. Expression
    18≡−1(mod19)18 \equiv -1 \pmod{19}
    Explanation

    Rewrite 18 as its negative residue modulo 19.

    Justification

    Congruence reduction modulo 19 shown on the board.

    Shown in the video
  2. Expression
    18−1≡1(mod19)18^{-1} \equiv 1 \pmod{19}
    Explanation

    Conclude that the inverse of 18 modulo 19 is 1.

    Justification

    The video presents this as the immediate consequence of the preceding congruence; the underlying fact is that -1 is its own inverse modulo 19.

    Shown in the video
Conclusion

18−1≡1(mod19)18^{-1} \equiv 1 \pmod{19}.

Derivation of 17−117^{-1} modulo 19

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board shows 17≡−2(mod19)17 \equiv -2 \pmod{19}, (-2)^{-1} ≡−10(mod19)\equiv -10 \pmod{19}, and 17−1≡9(mod19)17^{-1} \equiv 9 \pmod{19}.

  2. Audio
    Observation

    The speaker mentions the longer route ending at 17×9=15317 \times 9 = 153.

Uncertainties
  1. The clip does not show the intermediate verification 153≡1(mod19)153 \equiv 1 \pmod{19}; it only states 17×9=15317 \times 9 = 153 in the audio.

Proof
Steps
  1. Expression
    17≡−2(mod19)17 \equiv -2 \pmod{19}
    Explanation

    Replace 17 by the smaller negative residue -2 modulo 19.

    Justification

    Congruence reduction modulo 19 shown on the board.

    Shown in the video
  2. Expression
    (−2)−1≡−10(mod19)(-2)^{-1} \equiv -10 \pmod{19}
    Explanation

    Use the inverse of -2 modulo 19.

    Justification

    Written directly on the board as a known intermediate step.

    Shown in the video
  3. Expression
    17−1≡9(mod19)17^{-1} \equiv 9 \pmod{19}
    Explanation

    Convert -10 to the positive residue 9 modulo 19 and record the inverse of 17.

    Justification

    Since -10 ≡9(mod19)\equiv 9 \pmod{19}, the board rewrites the inverse in positive form.

    Shown in the video
Conclusion

17−1≡9(mod19)17^{-1} \equiv 9 \pmod{19}.

Why 5 has no inverse modulo 10

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board writes 1≡11≡21≡31≡1 \equiv 11 \equiv 21 \equiv 31 \equiv and then (5,10)=5 NOT 1 and No inverse.

  2. Audio
    Observation

    The speaker says adding 10 to 1 gives 11, 21, 31, and one would never get a multiple of 5.

  3. Audio
    Observation

    The speaker identifies the problem as gcd(5,10)=5, not 1.

Uncertainties
  1. The written congruence line is visually incomplete after the final ≡\equiv sign; the intended continuation is supplied by the audio as 11, 21, 31, ...

Proof
Steps
  1. Expression
    1≡11≡21≡31≡⋯(mod10)1 \equiv 11 \equiv 21 \equiv 31 \equiv \cdots \pmod{10}
    Explanation

    Begin listing numbers congruent to 1 modulo 10 by repeatedly adding 10.

    Justification

    The board shows 1≡11≡21≡31≡1 \equiv 11 \equiv 21 \equiv 31 \equiv, and the audio explains the pattern of adding 10.

    Shown in the video
  2. Expression
    No term in 1,11,21,31,… is a multiple of 5\text{No term in }1,11,21,31,\ldots\text{ is a multiple of }5
    Explanation

    Observe that this sequence never reaches a multiple of 5.

    Justification

    Stated explicitly in the audio: "you would never get a multiple of 5."

    Shown in the video
  3. Expression
    (5,10)=5≠1(5,10)=5 \ne 1
    Explanation

    Compute the greatest common divisor of 5 and 10.

    Justification

    Written on the board and spoken aloud.

    Shown in the video
  4. Expression
    No inverse\text{No inverse}
    Explanation

    Conclude that 5 has no inverse modulo 10.

    Justification

    The speaker states the general rule that gcd greater than 1 implies no inverse.

    Shown in the video
Conclusion

There is no inverse of 5 modulo 10 because (5,10)=5≠15 \ne 1.

Worked examples · 7

Example: Find the inverse of 5 mod 7

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Prompt on screen: “Find the inverse of 5 mod 7.”

  2. Formula
    Observation

    Work shown: “1 ≡ 8 ≡ 15 (mod 7)” with 15 split into 5 and 3.

  3. Formula
    Observation

    Conclusion written: “5−15^{-1} ≡ 3 (mod 7)”.

Problem

Find the inverse of 5 mod 7.

Given
  1. Target integer: 5.

  2. Modulus: 7.

Goal

Find an integer b such that 5b ≡ 1 (mod 7).

Steps
  1. Expression
    1≡8(mod7)1 \equiv 8 \pmod{7}
    Explanation

    Add 7 to 1 to search within the congruence class of 1 modulo 7.

    Justification

    Congruence is preserved when adding the modulus.

    Shown in the video
  2. Expression
    1≡8≡15(mod7)1 \equiv 8 \equiv 15 \pmod{7}
    Explanation

    Add 7 again to reach 15.

    Justification

    15 is still congruent to 1 modulo 7.

    Shown in the video
  3. Expression
    15=5⋅315 = 5 \cdot 3
    Explanation

    Recognize 15 as a multiple of 5.

    Justification

    Integer factorization.

    Shown in the video
  4. Expression
    5−1≡3(mod7)5^{-1} \equiv 3 \pmod{7}
    Explanation

    Conclude that 3 is the inverse of 5 modulo 7.

    Justification

    Because 5⋅35\cdot 3 ≡ 1 (mod 7).

    Shown in the video
Answer

5−1≡3(mod7)5^{-1} \equiv 3 \pmod{7}

Verification

5⋅3=155\cdot 3 = 15 and 15 ≡ 1 (mod 7).

Example: Find the inverse of 5 mod 11

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Prompt on screen: “Find the inverse of 5 mod 11.”

  2. Formula
    Observation

    Work shown: “1 ≡ 12 ≡ 23 ≡ 34 ≡ 45 (mod 11)” with 45 split into 5 and 9.

  3. Formula
    Observation

    Conclusion written: “5−15^{-1} ≡ 9 (mod 11)”.

Problem

Find the inverse of 5 mod 11.

Given
  1. Target integer: 5.

  2. Modulus: 11.

Goal

Find an integer b such that 5b ≡ 1 (mod 11).

Steps
  1. Expression
    1≡12(mod11)1 \equiv 12 \pmod{11}
    Explanation

    Start with 1 and add 11.

    Justification

    Adding the modulus preserves congruence modulo 11.

    Shown in the video
  2. Expression
    12≡23≡34≡45(mod11)12 \equiv 23 \equiv 34 \equiv 45 \pmod{11}
    Explanation

    Keep adding 11 until reaching a multiple of 5.

    Justification

    All listed numbers remain congruent to 1 modulo 11.

    Shown in the video
  3. Expression
    45=5⋅945 = 5 \cdot 9
    Explanation

    Factor 45 as 5 times 9.

    Justification

    Integer factorization.

    Shown in the video
  4. Expression
    5−1≡9(mod11)5^{-1} \equiv 9 \pmod{11}
    Explanation

    Conclude that 9 is the inverse of 5 modulo 11.

    Justification

    Because 5⋅95\cdot 9 ≡ 1 (mod 11).

    Shown in the video
Answer

5−1≡9(mod11)5^{-1} \equiv 9 \pmod{11}

Verification

5⋅9=455\cdot 9 = 45 and 45 ≡ 1 (mod 11).

Example: Find the inverse of 6 mod 17

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Prompt on screen: “Find the inverse of 6 mod 17.”

  2. Formula
    Observation

    Work shown: “1 ≡ 18 (mod 17)” with 18 split into 6 and 3.

  3. Formula
    Observation

    Conclusion written: “6−16^{-1} ≡ 3 (mod 17)”.

Problem

Find the inverse of 6 mod 17.

Given
  1. Target integer: 6.

  2. Modulus: 17.

Goal

Find an integer b such that 6b ≡ 1 (mod 17).

Steps
  1. Expression
    1≡18(mod17)1 \equiv 18 \pmod{17}
    Explanation

    Add 17 to 1 to get 18.

    Justification

    Adding the modulus preserves congruence modulo 17.

    Shown in the video
  2. Expression
    18=6⋅318 = 6 \cdot 3
    Explanation

    Factor 18 as 6 times 3.

    Justification

    Integer factorization.

    Shown in the video
  3. Expression
    6−1≡3(mod17)6^{-1} \equiv 3 \pmod{17}
    Explanation

    Conclude that 3 is the inverse of 6 modulo 17.

    Justification

    Because 6⋅36\cdot 3 ≡ 1 (mod 17).

    Shown in the video
Answer

6−1≡3(mod17)6^{-1} \equiv 3 \pmod{17}

Verification

6⋅3=186\cdot 3 = 18 and 18 ≡ 1 (mod 17).

Example: inverse of 18 modulo 19

Clear evidence
Shown in the video
Evidence
  1. Caption evidence
    Observation

    The typed prompt on screen reads 'Find the inverse of 18 mod 19.'

  2. Formula
    Observation

    The board solves it first by listing 1 ≡ 20 ≡ 39 ≡ 58 ≡ 77 ≡ ... ≡ 324 and then by writing 18 ≡ -1 (mod 19).

Problem

Find the inverse of 18 mod 19.

Given
  1. Modulus m=19m = 19.

  2. Residue a=18a = 18.

Goal

Find b such that 18b ≡ 1 (mod 19).

Steps
  1. Expression
    1≡20≡39≡58≡77≡⋯≡324(mod19)1 \equiv 20 \equiv 39 \equiv 58 \equiv 77 \equiv \cdots \equiv 324 \pmod{19}
    Explanation

    Generate numbers congruent to 1 modulo 19 by repeatedly adding 19.

    Justification

    Congruence is preserved when adding the modulus.

    Shown in the video
  2. Expression
    324=18⋅18324 = 18 \cdot 18
    Explanation

    Notice that 324 is divisible by 18, with quotient 18.

    Justification

    Factorization shown on the board.

    Shown in the video
  3. Expression
    18−1≡18(mod19)18^{-1} \equiv 18 \pmod{19}
    Explanation

    Conclude that 18 is its own inverse modulo 19.

    Justification

    Because 18⋅1818\cdot 18 ≡ 1 (mod 19).

    Shown in the video
  4. Expression
    18≡−1(mod19),(−1)2=118 \equiv -1 \pmod{19},\quad (-1)^2=1
    Explanation

    A faster route is to rewrite 18 as -1 and use that -1 is self-inverse.

    Justification

    Negative residues are congruent modulo 19, and (-1)(-1)=1.

    Shown in the video
Answer

18−118^{-1} ≡ 18 (mod 19).

Verification

Check: 18⋅18=32418\cdot 18 = 324 and 324 ≡ 1 (mod 19); equivalently, 18 ≡ -1 and (-1)^2 ≡ 1.

Example: inverse of 17 modulo 19

Clear evidence
Shown in the video
Evidence
  1. Caption evidence
    Observation

    The handwritten prompt reads 'Find inverse of 17 mod 19.'

  2. Formula
    Observation

    The board writes 17 ≡ -2 (mod 19), (-2)^{-1} ≡ -10 (mod 19), and 17−117^{-1} ≡ 9 (mod 19).

Problem

Find the inverse of 17 mod 19.

Given
  1. Modulus m=19m = 19.

  2. Residue a=17a = 17.

  3. Earlier in the clip, 1 ≡ 20 (mod 19) was already established.

Goal

Find b such that 17b ≡ 1 (mod 19).

Steps
  1. Expression
    17≡−2(mod19)17 \equiv -2 \pmod{19}
    Explanation

    Replace 17 by the simpler negative residue -2.

    Justification

    Adding 19 to -2 gives 17.

    Shown in the video
  2. Expression
    20=(−2)(−10)20 = (-2)(-10)
    Explanation

    Factor the previously noted number 20, which is congruent to 1 modulo 19.

    Justification

    Direct arithmetic identity.

    Shown in the video
  3. Expression
    (−2)−1≡−10(mod19)(-2)^{-1} \equiv -10 \pmod{19}
    Explanation

    Thus -10 is the inverse of -2 modulo 19.

    Justification

    Because (-2)(-10)=20 ≡ 1 (mod 19).

    Shown in the video
  4. Expression
    −10+19=9-10 + 19 = 9
    Explanation

    Convert the negative inverse to a positive representative.

    Justification

    Adding the modulus preserves congruence.

    Shown in the video
  5. Expression
    17−1≡9(mod19)17^{-1} \equiv 9 \pmod{19}
    Explanation

    Therefore the inverse of 17 modulo 19 is 9.

    Justification

    Substitute back 17 ≡ -2.

    Shown in the video
Answer

17−117^{-1} ≡ 9 (mod 19).

Verification

Check: 17⋅9=15317\cdot 9 = 153 and 153=8⋅19+1153 = 8\cdot 19 + 1, so 17⋅917\cdot 9 ≡ 1 (mod 19).

Find inverses of 18 and 17 modulo 19

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Printed prompt: "Find the inverse of 18 mod 19."

  2. Formula
    Observation

    Handwritten work includes 18≡−1(mod19)18 \equiv -1 \pmod{19}, 18−1≡1(mod19)18^{-1} \equiv 1 \pmod{19}, Find inverse of 17 mod 19, 17≡−2(mod19)17 \equiv -2 \pmod{19}, (-2)^{-1} ≡−10(mod19)\equiv -10 \pmod{19}, 17−1≡9(mod19)17^{-1} \equiv 9 \pmod{19}.

  3. Audio
    Observation

    The speaker says the long way would reach 17×9=15317 \times 9 = 153.

Uncertainties
  1. The top congruence chain visible on the board is only partially legible in this clip; the exact full sequence before 324 cannot be transcribed reliably from the provided frames.

Problem

Compute the modular inverses of 18 and 17 modulo 19.

Given
  1. Modulus m=19m=19.

  2. Numbers whose inverses are requested: 18 and 17.

  3. Intermediate congruences shown: 18≡−1(mod19)18 \equiv -1 \pmod{19}, 17≡−2(mod19)17 \equiv -2 \pmod{19}, (-2)^{-1} ≡−10(mod19)\equiv -10 \pmod{19}.

Goal

Express each inverse as a residue modulo 19.

Steps
  1. Expression
    18≡−1(mod19)18 \equiv -1 \pmod{19}
    Explanation

    Reduce 18 to a simpler representative modulo 19.

    Justification

    Shown on the board.

    Shown in the video
  2. Expression
    18−1≡1(mod19)18^{-1} \equiv 1 \pmod{19}
    Explanation

    Record the inverse of 18.

    Justification

    Directly written on the board as the next line.

    Shown in the video
  3. Expression
    17≡−2(mod19)17 \equiv -2 \pmod{19}
    Explanation

    Reduce 17 to a simpler representative modulo 19.

    Justification

    Shown on the board under "Find inverse of 17 mod 19."

    Shown in the video
  4. Expression
    (−2)−1≡−10(mod19)(-2)^{-1} \equiv -10 \pmod{19}
    Explanation

    Use the inverse of -2 modulo 19.

    Justification

    Shown on the board.

    Shown in the video
  5. Expression
    17−1≡9(mod19)17^{-1} \equiv 9 \pmod{19}
    Explanation

    Rewrite -10 as 9 modulo 19 to give the positive residue for the inverse of 17.

    Justification

    Shown on the board; the conversion uses -10 ≡9(mod19)\equiv 9 \pmod{19}.

    Shown in the video
Answer

18−1≡1(mod19)18^{-1} \equiv 1 \pmod{19} and 17−1≡9(mod19)17^{-1} \equiv 9 \pmod{19}.

Verification

The audio notes that the longer route for 17 reaches 17×9=15317 \times 9 = 153, consistent with checking the product before reducing modulo 19.

Find the inverse of 5 modulo 10

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Printed prompt: "Find the inverse of 5 mod 10."

  2. Formula
    Observation

    Handwritten work includes 1≡11≡21≡31≡1 \equiv 11 \equiv 21 \equiv 31 \equiv, (5,10)=5 NOT 1, and No inverse.

  3. Audio
    Observation

    The speaker says adding 10 repeatedly never yields a multiple of 5, then states gcd(5,10)=5, not 1, so there is no inverse.

Uncertainties
  1. The written congruence line ends with a trailing ≡\equiv sign without a visible next numeral; the audio supplies the intended continuing pattern.

Problem

Determine whether 5 has a multiplicative inverse modulo 10, and find it if it exists.

Given
  1. Modulus m=10m=10.

  2. Number whose inverse is requested: 5.

Goal

Decide existence of 5−1(mod10)5^{-1} \pmod{10} and conclude accordingly.

Steps
  1. Expression
    1≡11≡21≡31≡⋯(mod10)1 \equiv 11 \equiv 21 \equiv 31 \equiv \cdots \pmod{10}
    Explanation

    Try the direct search by listing numbers congruent to 1 modulo 10.

    Justification

    The board writes the beginning of this chain, and the audio explains adding 10 to 1, then 11, 21, 31.

    Shown in the video
  2. Expression
    never get a multiple of 5\text{never get a multiple of }5
    Explanation

    Notice that none of these candidates is divisible by 5, so no product 5x can equal 1 modulo 10 by this route.

    Justification

    Explicitly stated in the audio.

    Shown in the video
  3. Expression
    (5,10)=5(5,10)=5
    Explanation

    Compute the greatest common divisor of 5 and 10.

    Justification

    Written on the board and spoken aloud.

    Shown in the video
  4. Expression
    5≠15 \ne 1
    Explanation

    Compare the gcd with 1.

    Justification

    The board writes NOT 1 beside (5,10)=5.

    Shown in the video
  5. Expression
    No inverse\text{No inverse}
    Explanation

    Conclude that 5 has no inverse modulo 10.

    Justification

    Written on the board and justified by the spoken gcd rule.

    Shown in the video
Answer

No inverse.

Verification

The example is checked by the explicit gcd computation (5,10)=5≠15 \ne 1 and by the failed search through 1, 11, 21, 31, ...

Visual events · 10

Title slide with definition and theorem

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    A white slide titled “Modular Inverses” shows a definition line and a theorem line in black text.

  2. Animation
    Observation

    Purple underlines are drawn sequentially beneath key phrases and symbols in the definition and theorem.

Objects
  1. Title “Modular Inverses”.

  2. Definition sentence.

  3. Theorem sentence.

  4. Purple underline strokes.

Changes
  1. Underline appears under “multiplicative inverses modulo m” around 10s.

  2. Underline appears under “ab ≡ 1 (mod m)” around 12s–14s.

  3. Underline appears under “a” in the theorem around 25s.

  4. Underline appears under “modulo m” around 26s–27s.

  5. Underline appears under “(a,m)=1” around 28s–29s.

Invariants
  1. The slide text itself remains unchanged while annotations are added.

Interpretation

The visual emphasis tracks the spoken explanation, highlighting the definition of inverse and the gcd condition in the theorem.

Worked example for inverse of 5 modulo 7

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    New slide prompt: “Find the inverse of 5 mod 7.”

  2. Animation
    Observation

    Purple handwriting builds the congruence chain and factor tree step by step.

Objects
  1. Prompt text.

  2. Handwritten congruence chain “1 ≡ 8 ≡ 15 (mod 7)”.

  3. Factor branches from 15 to 5 and 3.

  4. Final notation “5−15^{-1} ≡ 3 (mod 7)”.

Changes
  1. First “1 ≡ 8” is written.

  2. Then “≡ 15” is appended.

  3. Then 15 is factored into 5 and 3 below.

  4. Finally the inverse statement is written underneath.

Invariants
  1. The modulus remains 7 throughout the example.

Interpretation

The animation shows the search process: move through numbers congruent to 1 modulo 7 until a multiple of 5 is found, then read off the inverse.

Worked example for inverse of 5 modulo 11

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    New slide prompt: “Find the inverse of 5 mod 11.”

  2. Animation
    Observation

    Purple handwriting extends a longer congruence chain and then factors the terminal number.

Objects
  1. Prompt text.

  2. Handwritten chain “1 ≡ 12 ≡ 23 ≡ 34 ≡ 45 (mod 11)”.

  3. Factor branches from 45 to 5 and 9.

  4. Final notation “5−15^{-1} ≡ 9 (mod 11)”.

Changes
  1. The chain grows term by term from 12 to 23 to 34 to 45.

  2. After 45 is reached, it is factored into 5 and 9.

  3. The inverse statement is then written below.

Invariants
  1. The modulus remains 11 throughout the example.

Interpretation

This visual sequence demonstrates the same add-the-modulus search method with more iterations before hitting a multiple of 5.

Worked example for inverse of 6 modulo 17

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    New slide prompt: “Find the inverse of 6 mod 17.”

  2. Animation
    Observation

    Purple handwriting shows “1 ≡ 18 (mod 17)” and factors 18 into 6 and 3.

Objects
  1. Prompt text.

  2. Handwritten congruence “1 ≡ 18 (mod 17)”.

  3. Factor branches from 18 to 6 and 3.

  4. Final notation “6−16^{-1} ≡ 3 (mod 17)”.

Changes
  1. 18 is written as the next representative after 1.

  2. 18 is factored into 6 and 3.

  3. The inverse statement is written below.

Invariants
  1. The modulus remains 17 throughout the example.

Interpretation

The shortest example shows that sometimes one addition of the modulus already yields a multiple of the target integer.

Transition to the new problem

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    The screen changes from a previous problem about 6 mod 17 to a new typed prompt 'Find the inverse of 18 mod 19.'

Objects
  1. Previous handwritten solution for 6 mod 17

  2. New typed prompt for 18 mod 19

Changes
  1. Old work disappears.

  2. New problem statement appears at the top.

Invariants
  1. The interface remains a digital whiteboard with purple handwriting.

Interpretation

This marks the start of the main example in the clip.

Writing the congruence chain and factorization

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    Purple handwriting successively adds terms to the chain 1 ≡ 20 ≡ 39 ≡ 58 ≡ 77 ≡ ... ≡ 324, then writes the factor pair 18 and 18 beneath 324.

Objects
  1. Congruence chain beginning with 1

  2. Number 324

  3. Factor labels 18 and 18

Changes
  1. Each new term is obtained by adding 19.

  2. The chain ends at 324.

  3. 324 is decomposed visually into 18⋅1818\cdot 18.

Invariants
  1. All displayed terms remain in the same congruence class as 1 modulo 19.

Interpretation

The visual process demonstrates a search for a multiple of 18 inside the congruence class of 1 modulo 19.

Rewriting 18 as -1 modulo 19

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    Below the earlier work, the presenter writes 18 ≡ -1 (mod 19), then rewrites the inverse conclusion as 18−118^{-1} ≡ 18 (mod 19).

Objects
  1. Expression 18 ≡ -1 (mod 19)

  2. Expression 18−118^{-1} ≡ 18 (mod 19)

Changes
  1. The positive residue 18 is replaced by the negative residue -1.

  2. The inverse statement is restated using the shortcut.

Invariants
  1. The modulus stays 19.

  2. The residue class represented by 18 does not change.

Interpretation

The visual rewrite emphasizes that choosing a convenient representative can simplify inverse calculations.

Second example written below a divider

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    A horizontal divider is drawn, then the new prompt 'Find inverse of 17 mod 19' is written, followed by 17 ≡ -2 (mod 19), (-2)^{-1} ≡ -10 (mod 19), and 17−117^{-1} ≡ 9 (mod 19).

Objects
  1. Divider line

  2. Prompt 'Find inverse of 17 mod 19'

  3. Congruences involving -2, -10, and 9

Changes
  1. The board shifts from the first example to a second example.

  2. The solution proceeds through negative residues and then converts back to a positive answer.

Invariants
  1. The modulus remains 19 throughout.

Interpretation

The layout separates two applications of the same shortcut idea.

Transition from the modulo-19 example to the modulo-10 example

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    The display changes from the completed modulo-19 board to a new page headed "Find the inverse of 5 mod 10."

Objects
  1. Completed handwritten work for 18 and 17 modulo 19

  2. New printed heading "Find the inverse of 5 mod 10."

Changes
  1. The previous board disappears.

  2. A fresh whiteboard page appears with only the new problem statement at first.

Invariants
  1. The lesson remains focused on finding modular inverses.

  2. The notation style stays the same, using handwritten purple math on a digital whiteboard.

Interpretation

This visual transition marks a shift from successful inverse computations to a counterexample showing when an inverse does not exist.

Writing the gcd obstruction and conclusion

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    The presenter writes (5,10)=5, then adds NOT 1, then writes No inverse underneath.

Objects
  1. Expression (5,10)=5

  2. Annotation NOT 1

  3. Conclusion No inverse

Changes
  1. First the gcd value is written.

  2. Then the comparison to 1 is added.

  3. Finally the nonexistence conclusion is written below.

Invariants
  1. The modulus remains 10 and the number remains 5 throughout this sequence.

Interpretation

The visual order encodes the reasoning: compute the gcd, observe it is not 1, and therefore conclude that no modular inverse exists.

Misconceptions · 5

Assuming an inverse exists even when gcd(a,m)>1

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker explicitly says, “If the greatest common divisor between a and m is greater than 1, then there is no inverse mod m.”

Misconception

One might think any integer a has some inverse modulo m regardless of its common factors with m.

Clarification

The video states that if gcd(a,m)>1, then a has no inverse modulo m.

Brute force is valid but inefficient

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says the brute-force method will always work, but sometimes you end up spending a lot of time trying to find the inverse.

Misconception

One might think that repeatedly adding the modulus until a multiple appears is the best general strategy.

Clarification

The clip explicitly contrasts this reliable but slow method with a faster shortcut using negative residues.

Negative residues are legitimate intermediate forms

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says we are used to adding 19, but we can subtract 19 as well, and later adds 19 back to -10 because they want positive numbers.

  2. Formula
    Observation

    The board uses 18 ≡ -1 (mod 19), 17 ≡ -2 (mod 19), and (-2)^{-1} ≡ -10 (mod 19).

Misconception

Students may believe modular arithmetic must always stay within positive representatives.

Clarification

The video shows that negative residues are valid and often simpler, though one may convert back to a positive representative at the end.

Assuming every integer has an inverse modulo any modulus

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker contrasts the earlier successful examples with "what if I was asked to find the inverse of 5 mod 10?" and then shows that no inverse exists.

  2. Formula
    Observation

    The board writes No inverse after (5,10)=5 NOT 1.

Misconception

One might expect the same search method to always produce an inverse, as it did in the modulo-19 examples.

Clarification

The video shows that this fails for 5 mod 10: the search through numbers congruent to 1 modulo 10 never reaches a multiple of 5, and the gcd test explains why no inverse exists.

Thinking brute-force listing is the only practical method

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    At the start the speaker says the long way would eventually get to 17×9=15317 \times 9 = 153, "it just takes a little bit longer."

  2. Audio
    Observation

    Later the speaker says one could "spend the rest of your life adding 10" and still never get a multiple of 5.

Misconception

A learner may think modular inverses must always be found by extending a long list of congruences.

Clarification

The clip demonstrates faster structural methods: rewrite numbers as smaller residues modulo the modulus, and check gcd(a,m) to decide existence immediately.

Concept relations · 11

Definition of multiplicative inverses modulo m → Existence criterion for a modular inverse

Clear evidence
Derived from the video
Evidence
  1. Formula
    Observation

    The theorem directly refers to “has an inverse modulo m,” which is defined in the preceding definition line.

Prerequisite
Explanation

The existence theorem is about the notion introduced in the definition of multiplicative inverses modulo m.

Existence criterion for a modular inverse → Search method for modular inverses by adding the modulus

Clear evidence
Derived from the video
Evidence
  1. Audio
    Observation

    After stating the theorem, the speaker moves to examples and searches for actual inverses.

  2. Formula
    Observation

    Each example finds a concrete inverse consistent with the theorem’s existence condition.

Application
Explanation

The worked examples apply the existence criterion by finding inverses in cases where the integers are coprime to the modulus.

Search method for modular inverses by adding the modulus → Example: Find the inverse of 5 mod 7

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    All three examples use the same pattern of writing 1 ≡ 1+m1+m ≡ 1+2m1+2m ≡ … until a multiple of the target integer appears.

Application
Explanation

The first example is a direct application of the add-the-modulus search method.

Search method for modular inverses by adding the modulus → Example: Find the inverse of 5 mod 11

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The second example repeats the same search pattern with modulus 11.

Application
Explanation

The second example applies the same method with a longer search chain.

Search method for modular inverses by adding the modulus → Example: Find the inverse of 6 mod 17

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The third example again starts from 1 and adds the modulus once to reach a multiple of the target integer.

Application
Explanation

The third example applies the same method in the simplest case, requiring only one addition of the modulus.

Modular multiplicative inverse → Brute-force search by adding the modulus

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The search for a number congruent to 1 that is divisible by 18 directly implements the definition of inverse.

Application
Explanation

The brute-force method is a direct application of the definition ab ≡ 1 (mod m).

Brute-force search by adding the modulus → Use negative residues to simplify inverse problems

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    After noting the slow method, the speaker introduces 'a little bit of a shortcut.'

Contrast
Explanation

The shortcut is presented as a faster alternative to the exhaustive search.

Use negative residues to simplify inverse problems → The residue -1 is its own inverse

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The shortcut is applied by rewriting 18 as -1 and then using (-1)^2=12=1.

Special case
Explanation

The fact that -1 is self-inverse is a special case enabled by replacing a residue with a negative representative.

Use negative residues to simplify inverse problems → Convert a negative inverse back to a positive residue

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    After obtaining (-2)^{-1} ≡ -10, the board converts to 9 by adding 19.

Application
Explanation

Once a negative-residue inverse is found, the final step is to express it as a positive residue.

Criterion for nonexistence of a modular inverse → Meaning of a modular inverse in this lesson

Clear evidence
Derived from the video
Evidence
  1. Audio
    Observation

    The second example first attempts a search, then stops and uses the gcd to explain the failure.

  2. Formula
    Observation

    The board moves from the attempted congruence chain to (5,10)=5 NOT 1 and No inverse.

Uncertainties
  1. This ordering is inferred from the presentation flow rather than stated as an explicit theorem.

Contrast
Explanation

The lesson contrasts cases where an inverse can be computed with a case where the inverse does not exist at all.

Find the inverse of 5 modulo 10 → If gcd(a,m)>1, then a has no inverse modulo m

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    After the specific computation (5,10)=5, the speaker generalizes: "if there's a greatest common divisor greater than 1 between those two values, then there is no inverse."

Application
Explanation

The specific example of 5 modulo 10 is used to motivate and illustrate the general nonexistence rule.

Find an answer · 15

What does it mean for two integers to be multiplicative inverses modulo m?

Clear evidence
Derived from the video
Evidence
  1. Formula
    Observation

    The definition line is visible on the title slide.

Knowledge points
  1. Definition of multiplicative inverses modulo m

When does an integer have an inverse modulo m?

Clear evidence
Derived from the video
Evidence
  1. Formula
    Observation

    The theorem line states the gcd condition.

Knowledge points
  1. Existence criterion for a modular inverse

Why is there no inverse modulo m if gcd(a,m) is greater than 1?

Clear evidence
Derived from the video
Evidence
  1. Audio
    Observation

    The speaker states the case gcd(a,m)>1 gives no inverse.

Knowledge points
  1. Existence criterion for a modular inverse
  2. Assuming an inverse exists even when gcd(a,m)>1

How can I find a modular inverse by hand using the add-the-modulus method?

Clear evidence
Derived from the video
Evidence
  1. Formula
    Observation

    The examples all show the same handwritten search procedure.

Knowledge points
  1. Search method for modular inverses by adding the modulus
  2. Example: Find the inverse of 5 mod 7
  3. Example: Find the inverse of 5 mod 11
  4. Example: Find the inverse of 6 mod 17

What is the inverse of 5 modulo 7 and how is it computed?

Clear evidence
Derived from the video
Evidence
  1. Formula
    Observation

    The first worked example is fully shown on screen.

Knowledge points
  1. Example: Find the inverse of 5 mod 7

What is the inverse of 5 modulo 11 and how is it computed?

Clear evidence
Derived from the video
Evidence
  1. Formula
    Observation

    The second worked example is fully shown on screen.

Knowledge points
  1. Example: Find the inverse of 5 mod 11

What is the inverse of 6 modulo 17 and how is it computed?

Clear evidence
Derived from the video
Evidence
  1. Formula
    Observation

    The third worked example is fully shown on screen.

Knowledge points
  1. Example: Find the inverse of 6 mod 17

Why is 18 its own inverse modulo 19?

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board shows 18 ≡ -1 (mod 19) and concludes 18−118^{-1} ≡ 18 (mod 19).

Knowledge points
  1. Modular multiplicative inverse
  2. Use negative residues to simplify inverse problems
  3. The residue -1 is its own inverse

How do negative residues make finding modular inverses easier?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker explains that we can subtract 19 as well as add it.

  2. Formula
    Observation

    The examples rewrite 18 as -1 and 17 as -2.

Knowledge points
  1. Use negative residues to simplify inverse problems
  2. Convert a negative inverse back to a positive residue

What is the inverse of 17 modulo 19 and how is it computed quickly?

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board derives 17−117^{-1} ≡ 9 (mod 19).

Knowledge points
  1. Use negative residues to simplify inverse problems
  2. Convert a negative inverse back to a positive residue

When should I use the brute-force congruence chain instead of the negative-residue shortcut?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker says the long method always works but can take a lot of time, then introduces a shortcut.

Knowledge points
  1. Brute-force search by adding the modulus
  2. Use negative residues to simplify inverse problems

Why does 5 have no inverse modulo 10?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The speaker asks what happens for the inverse of 5 mod 10 and then explains the gcd obstruction.

  2. Formula
    Observation

    Board text: (5,10)=5 NOT 1, No inverse.

Knowledge points
  1. Find the inverse of 5 modulo 10
  2. Criterion for nonexistence of a modular inverse
  3. If gcd(a,m)>1, then a has no inverse modulo m
Coverage and review notes

Covered · Title slide presents the definition of multiplicative inverses modulo m and the theorem giving the gcd existence criterion, with spoken explanation and purple annotations.

Covered · First worked example finds the inverse of 5 modulo 7 by adding the modulus until reaching a multiple of 5.

Covered · Second worked example finds the inverse of 5 modulo 11 using the same search method with a longer congruence chain.

Covered · Third worked example finds the inverse of 6 modulo 17, showing the shortest case of the method.

Covered · Transition from previous problem to the new prompt 'Find the inverse of 18 mod 19.'

Covered · First example solved by listing numbers congruent to 1 modulo 19 until reaching 324=18⋅18324 = 18\cdot 18.

Covered · The presenter introduces the shortcut 18 ≡ -1 (mod 19) and concludes 18 is its own inverse.

Covered · Second example finds the inverse of 17 modulo 19 via 17 ≡ -2 and (-2)^{-1} ≡ -10, then converts to 9.

Covered · Final hold on the completed board with no new mathematical content.

Covered · Completed modulo-19 examples for 18 and 17, including the spoken remark about the longer route reaching 17×9=15317 \times 9 = 153.

Covered · New problem "Find the inverse of 5 mod 10" and the attempted search 1≡11≡21≡31≡⋯1 \equiv 11 \equiv 21 \equiv 31 \equiv \cdots, with audio explaining that this never yields a multiple of 5.

Covered · Computation (5,10)=5 NOT 1, conclusion No inverse, and the generalized spoken rule about gcd greater than 1.

Covered · Closing remarks only: the speaker says the lesson should help, suggests practice, and ends with "have a great day"; no additional mathematical content appears.

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  • Modular inverses ExplanationAt 0:23
    Why this connection?

    Reviewed English material v1: card "Existence theorem for modular inverses" at 0:23 — An integer a has an inverse modulo m if and only if gcd(a,m)=1. If the gcd is greater than 1, no inverse exists.