2

I know how to calculate Modular Multiplicative Inverse. But I am unable to understand the physical meaning (Analytical Meaning/Intuitive Meaning ) of Modular Multiplicative Inverse. For Example

If we have to find the Multiplicative Inverse of 7 mod 5 then we know that it is 3. But If we think analytically and forget about the formula and Congruent type of things and use Basic Mathematics.

Then we get that the Answer is 1/2 because According we have to find the Multiplicative Inverse of 7 mod 5 and 7 mod 5 = 2 and the inverse of 2 is 1/2.

Why there is a big difference between both of these Answer. It may be a silly question but I am highly confused

Is there any difference between Normal Multiplicative Inverse and Modular Multiplicative Inverse?

Bhaskar
  • 177
  • Related: http://math.stackexchange.com/questions/1030943/is-modular-arithmetic-defined-for-all-rational-numbers-with-denominators-coprim/1898190#1898190 – Ethan Bolker Aug 20 '16 at 23:45
  • 3
    The inverse of 2 is 3 if we're working mod 5. – littleO Aug 20 '16 at 23:47
  • 1
    In my book, if you’re working in the integers modulo $5$, then $1/2$ is $3$. Isn’t the product of $2$ and $3$ equal to $1$ there? – Lubin Aug 21 '16 at 04:43

2 Answers2

1

Let us first do additive modular arithmetic. This system has only the following elements. $\{0,1,2,3,4\}$. But note that $4+1=0\pmod 5$ in this system. So $1$ works like $-4$ of familiar all integers (infinite) number system.

Also in this system non-zero elements $\{1,2,3,4\}$ is closed for multiplication modulo 5. And we see $2\times3$ is $1\pmod 5$ So $3$ behaves like $\frac12$ of our familiar rational number system. So we can interpret $3$ as $1/2$ assuming multiplication is interpreted as mod 5.

If the time is 9pm now 23 hours from now it would be 8pm. To find this answer very few people would add 23 to 9. Most people subtract 1 from 9. That is because in mod 24 number system $23$ is $-1$.

P Vanchinathan
  • 19,504
0

There is a difference. Not only a small difference. You are working with two different multiplications here. Let $p$ be a prime number. Then your two multiplication operations are

$\cdot: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}, (x,y) \mapsto x\cdot y$

$\cdot_p: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}, (x,y) \mapsto (x\cdot y) \text{ mod } p$

Both are different operations, so both have different inverse elements. Note, that the existence of an inverse element $x^{-1}$ for every $x \in \mathbb{Z}, x \neq 0$ is not a trivial fact. For the $\cdot_p$ operation there exist inverse elements because $p$ is a prime number. This is one of the many wonderful properties of prime numbers.

When you determine the inverse elements for operation $\cdot$, you need to extend your set of numbers to the rational numbers $\mathbb{Q}$. But when you determine the inverse element of $\cdot_p$ for a $x \in \mathbb{Z}, x \neq 0$, you will always find it within $\{1, ..., p\} \subset \mathbb{Z}$, so you don't need to extend your numbers.

S. M. Roch
  • 941
  • Can you explain it in more detail as you are understanding a person who don't know Group Theory – Bhaskar Aug 21 '16 at 01:57
  • I can. Though the downvote isn't fair as I can't know your level of preknowledge. If this answer now is what you expect, please flag it as the answer. Otherwise clarify your question. – S. M. Roch Aug 21 '16 at 03:19