Multiplicative inverse

Question

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: $$14=7^{-1}mod11$$

Answer 1

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.

Answer 2

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:

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

Answer 3

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.