Inverse modular multiplicative

Question

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 $\frac{1}{4}$ ?

Answer 1

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

Answer 2

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.

Answer 3

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.

Answer 4

$\frac14\cong4\pmod 5$. 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.

Answer 5

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 $4(3) =12 \equiv 2 \mod (5)$

Answer 6

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.