# Inverse modular multiplicative

So this seems really confusing for me. Supposedly, 4-1 mod 5 = 4.

Isn't the inverse multiplicative of 4 equal to $$\frac{1}{4}$$?

If so shouldn't $$\frac{1}{4}$$ mod 5 be equal to $$\frac{1}{4}$$ ?

• It depends on what you mean by $\frac{1}{4}$. If you mean the rational number $\frac{1}{4}$, i.e. the decimal number $0.25$, no, this is not an element of $\Bbb Z_5$. If by $\frac{1}{4}$ you mean the multiplicative inverse of $4$, then yes $\frac{1}{4}\equiv 4\equiv -1\equiv 9\equiv 14 \equiv 19\equiv \cdots$ are different ways to represent the same thing in $\Bbb Z_5$. – JMoravitz Nov 5 '18 at 3:35
• "Isn't the inverse multiplicative of 4 equal to 14? " $\frac 14$ is not in $\mathbb Z_5$ at all so we cant say the multiplicative inverse is $\frac 14$. We need to solve $x$ so that $4 \times x \equiv 1 \pmod 5$. As $4 \times 4 = 16 \equiv 1 \pmod 5$ the value that $x$ can be is $4$. So – fleablood Nov 5 '18 at 4:17

What is $$1000-999$$? of course it is equal to $$1000-999$$ but we can also say that $$1000-999=1$$.

Similar reasoning, we have

$$4^{-1} \equiv 4^{-1} \pmod{5}$$

but can we find a $$y \in \{0,1,2,3,4\}$$ such that

$$4^{-1} \equiv y \pmod{5}$$

The answer is $$4$$ because $$4(4)=16 = 3(5)+1 \equiv 1 \pmod{5}$$.

Alternatively, $$4^{-1}\equiv (-1)^{-1} \equiv -1 \equiv 4 \pmod{5}$$.

I think it is a bit dangerous using $${^{-1}}$$ here. In words, you're looking for the multplicative inverse of $$[4]$$ in $$\Bbb Z_5$$. Since $$\gcd(4,5)=1$$, you may write $$1$$ as a combination of $$4$$ and $$5$$, as follows: $$1\cdot 5 + (-1)\cdot4 = 1.$$Reducing mod $$5$$ we get $$[1] = [1]\cdot\require{cancel}\cancelto{0}{[5]}+{\color{blue}{[-1]}}\cdot[4] = \color{blue}{[4]}\cdot[4].$$So, the multiplicative inverse of $$[4]$$ is $$[4]$$ itself.

Yes, it is the case that $$1/4=4 \mod (5)$$ because $$4(4)=16 \equiv 1 \mod (5)$$

Since $$5$$ is a prime number every thing works fine with multiplication and inverses.

You may even define fractions $$\mod (5)$$ and they behave very nicely.

For example $$1/3 = 2 \mod (5)$$ because $$2(3) =6 \equiv 1 \mod (5)$$

Now you have $$(1/3) + (1/3) = (2/3) \mod (5)$$ which says $$2+2=4 \mod (5)$$

Note that $$2/3 = 4 \mod (5)$$ because $$4(3) =12 \equiv 2 \mod (5)$$

$$\frac14\cong4\pmod 5$$. Because, multiplying by $$4$$, we get $$1\cong 16\pmod 5$$.

Or, $$4\cdot 4=16\cong1\pmod 5$$. So indeed $$4^{-1}\cong4\pmod 5$$.

This reasoning works if, as in this case, $$\operatorname{gcd}(4,5)=1$$. If not we could have more than one possible inverse.

When we are doing modular arithmetic $$\mod 5$$ then Our universe consists of five elements.

They are $$0$$ which represents all integers with remainder $$0$$ when divided by $$5$$.

And $$1$$ which represents all integers with remainder $$1$$ when divided by $$5$$.

And $$2$$ which... well, you get the idea.

But the point is $$0, 1, 2 ,3 ,4$$ are the only things that exist. Nothing else exists.

We have two operations $$+$$ and $$\times$$. The result $$a + b$$ is the element $$0, 1,2,3,4$$ in which the result one of the integers in $$a$$ plus one of the integers in $$b$$ is in. ANd ther result $$a \times b$$ is the element $$0, 1,2,3,4$$ in which one of the terms of $$a$$ multiplied by one of the terms of $$b$$ is in.

We have terms $$-a$$ these are the terms where $$a + (-a) = 0$$

Now, remember, the ONLY five things that exist in the ENTIRE universe are $$0, 1,2,3,4$$ and $$0+0 = 0$$ so $$-0 = 0$$ And $$1 + 4 = 0$$ so $$-1 = 4$$ and $$2 +3 = 0$$ so $$-2 = 3$$ and $$3+2 = 0$$ so $$-3 =2$$ and $$4 + 1 = 0$$ so $$-4 = 1$$.

Likewise we have a term we can write as $$\frac 1a$$ or $$a^{-1}$$. These are the values where $$a \times \frac 1a = 1$$.

Now again, there are only !!FIVE!! count them $$5$$ things in the universe. $$0 \times a = 1$$ never happens. So $$\frac 10$$ does not exist.

$$1 \times 1 = 1$$ so $$\frac 11 = 1$$.

$$2 \times 3 = 1$$ os $$\frac 12 = 3$$.

$$3\times 2 = 1$$ so $$\frac 13 = 2$$.

$$4 \times 4 = 1$$ so $$\frac 14 = 4$$.

Those are what $$\frac 1a$$ must equal because they are the only things that exist in the unviverse (which again, contians how many things? $$5$$ there are exactly $$5$$ things in the entire universe) so that $$a \times ??? = 1$$.

That's all there is too it.

In modular arithmetic, we only deal with integers. So there is no "$$\frac{1}{4}$$". However, in certain cases, we do have an INVERSE. That is an integer $$a^{-1}$$ such that $$aa^{-1}$$(mod n) = 1. Recall that means $$aa^{-1}$$ = 1 + $$kn$$ for some integer k.

In your example, you have that 4 $$\cdot$$ 4 = 16 = 1 + 15 = 1+ 3(5). Thus the inverse of 4 mod 5 = 4. Note that given different n, a$$^{-1}$$ will be different or even nonexistent.

• Best answer here! You get to the heart of what seemed to be confusing the asker. +1 – Sambo Nov 5 '18 at 3:28