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What is the multiplicative inverse of 7 modulo 11?

Is this correct: $$7 = 11(0) +7$$ $$11 = 7(1) +4$$ $$7 = 4(1) +3$$ $$4 = 3(1) +1$$

We then take 3 equations:

  1. $$4 = 11 + 7(-1)$$
  2. $$3 = 7 + 4(-1)$$
  3. $$1 = 4 + 3(-1)$$

We then use the 3 equations and replace them in each other like so:

3 and 2 $$4+(7+4(-1))(-1) = 1$$ which gives us $$ (2)4+7(-1)=1$$ 1 and our equation above $$(2)(11+7(-1))+7(-1) =1$$ This will then simplify to: $$11(2)+7(-3)=1$$ We then can say (we know that $11-(-3)=14$) $$11(2)+7(14) = 1mod 11$$ And we know that $11(2)$ goes to 0 so: $$14=7^{-1}mod11$$

Marc van Leeuwen
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Will
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3 Answers3

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No, at one step you should have $11+(-3)=8$, instead of $11-(-3)=14$. So $8=7^{-1}\pmod{11}$
And that fits because $8\times7=56=55+1=1\pmod{11}$
All the rest of your work was correct.

Empy2
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  • I have a quick question, why wouldn't we be able to apply this formula and calculate the multiplicative inverse of 6 modulo 8? – Will Oct 27 '14 at 13:28
  • Try it. You will end up with 2 instead of 1 – Empy2 Oct 27 '14 at 14:00
  • I can't seem to get anything. Is it impossible or non-exsistant? – Will Oct 27 '14 at 14:09
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    Yes, it is. All multiples of 6 have a factor 2, and that stays after you subtract mod 8. You never end up at 1, so 6 has no inverse. – Empy2 Oct 27 '14 at 14:26
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This will then simplify to:

$11(2)+7(−3)=1$

We then can say (we know that $-3=8\ mod11$)

$11(2)+7(8)=1mod11$

And we know that 11(2) goes to 0 so:

$8=7^{-1}\mod11$

jiten
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Leonardo Torres
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    @jiten Please do not make trivial edits to six year old questions and answers. That puts them on the active questions queue and wastes time for folks like me who pay attention to those questions. – Ethan Bolker Aug 10 '22 at 14:20
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I find it a bit strange that you've got 14 as the multiplicative inverse. We usually think of integers modulo some number $n$ under multiplication as the group $(\mathbb{Z}/n\mathbb{Z})^{\times}$, in which case we don't have 'numbers' (they're actually classes) bigger than $n$. (You've also got a nice case, as 11 is prime. If it were a composite number, $m$, the group would have to consist of classes $\bar n$ such that $\gcd(n,m)=1$, to satisfy the group axioms - failing this condition, some elements will not have inverses).

11 isn't very big, so case checking is the easiest way forward. If I'm looking for the inverse of $7\mod 11$ then I want some number $a\in[1,10]$ such that $7a=11b+1$ for some $b\in\mathbb{Z}$. Equivalently, $\dfrac{7a-1}{b}=11$. So I look at the first 10 multiples of 7 and see which one that is one greater than a multiple of 11. $7\times8=56$ is just what I want.

FireGarden
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