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How can we intuitively understand modulo multiplicative inverse?

Suppose we have an ring $mathbbZ_13= 0, 1, 2, 3, ldots, 12$.

Each element except zero has corresponding multiplicative inverse.

Below is the mapping of inverse.

1 $rightarrow$ 1

2 $rightarrow$ 7

3 $rightarrow$ 9

4 $rightarrow$ 10

5 $rightarrow$ 8

6 $rightarrow$ 11

7 $rightarrow$ 2

8 $rightarrow$ 5

9 $rightarrow$ 3

10 $rightarrow$ 4

11 $rightarrow$ 6

12 $rightarrow$ 12

Now, I want to consider division by 2, which here means that multiplication by 7.
However, some integer multiplied 7 becomes acutally the same result as division by 2 over real number field.
For example, the following holds.

4 * 7 = 28 = 2 mod 13;
6 * 7 = 42 = 3 mod 13;
etc.

On the other hand, other values do not produce the same result over real number fields.

For example,
5 * 7 = 35 = 9 mod 13 (I want this value to be 2 or 3 since 5/2 = 2.5)

7 * 7 = 49 = 10 mod 13 (I want this value to be 3 or 4 since 7/2 = 3.5).

Why some values produce the same result as over the real number field, and the others do not??

Operations mod $n$ aren't guaranteed to preserve the order inherited from the reals.

So the fact that in $mathbbR$, we have
$$2 < smallfrac52 < 3$$
doesn't imply
$$2 < (smallfrac52;textmod;13) < 3$$
However what is true is that, working mod $13$, we have
$$
smallfrac52 = 2+smallfrac12 = 2+7 = 9
qquad;;;
$$
and also
$$smallfrac52 = 3-smallfrac12 = 3-7 = -4 = 9$$

If $b$ is invertible mod $c$,
$ab^-1 equiv c mod p$ is equivalent to $a equiv b c mod p$, and this means $a = b c + k p$ for some integer $k$. Sometimes you'll have $k = 0$, so $a = b c$ and $c = a/b$ (as integers), sometimes you won't.

When your modulus is a prime number like $13$ you actually have a field and you may define fractions within your field.

For example $7=1/2$ and $10=1/4$
If you multiply $7times 7=49=10$ you notice that it is the same as $$1/2 times 1/2 =1/4$$

The problem which puzzles you is confusing the field of equivalency classes with the field of real numbers.

In the field of equivalency classes you have $$5/2 =5times 7=9$$

Now if you multiply $9times 2$ you get $18$ which is $5$ as it should be.

Note that $2.5$ makes sense in real field but you do not have a class of $2.5$ mod $13$ instead you have class of $9$ which serves well as $5/2$

Because that doesn't work even for the number $2$. Note that $11times7=77equiv12pmod13$. However, $frac112=5.5$, which is not near $12$.

Note that $5 times 7 = 35 = 9 pmod13$, not 11 as you claimed in your question -- and, given this, it might help to observe that we can also think of $2.5$ as $2.5 = 2 + .5 = 2 + 2^-1 = 2 + 7 = 9 pmod13$.

$$-12to -6to 10to -8to -4 to -2to 12to 6to -10to 8to 4to 2tocdots$$ is the looping structure you want here. All I did was turn odd elements into even equivalents. note how 0 never appears. That's because it's its own multiplicative inverse mod anything, except when division by 0 or zero divisors occurs.

Generally you can turn into equivalents and get the values. All equivalents that are 1 mod 3 have multiplicative inverse on division by 3 because 13 is 1 mod 3, so the difference is 0 mod 3 by mod addition rules. There aren't always equivalents that work mod composites.