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General question question in Linear Algebra

Hi guys, in my LA lecture yesterday the lecturer kept bringing forth addition problems that contradicted what i know as right. for example in F5(field of 5) the inverse of 4 is 4 and 3 4=2. Anyone care to explain what's going on?

February 18 2015, 9:04 am

ALL ANSWERS
**Houston Muzamhindo**
(University of Cape Town, South Africa (ZA))

Hi Boitumelo.

I'll explain the basics now and then when you understand that part, we can go to the harder stuff. Basically what he is talking about it called modular arithmetic. It has got to do with divisions and keeping reminders.

When they say "4 modulo 3" or "4 mod 3", they mean divide 4 by 3 and write what the remainder is, in this case the answer is 1. "5 mod 2 = 3" and "6 mod 10 = 6".

So in F5 (field of 5), the modulo or mod is 5. So we are looking for the inverse of 4 in mod 5. The simplest definition of inverse is:

"An inverse of x is a number such that when you multiply x with it, you will get the number 1, for example the inverse of 2 is 1/2 because when you multiply the 2 and 1/2 you will get 1."

That is just a basic definition. It becomes a little bit complicated, not too much lol, when it comes to modular arithmetic. The easiest way to find an inverse mod something is to ask yourself "What number should I multiply x with such that when I divide by the mod, I get 1". Let's do your example:

What number should I multiple 4 with so that when I divide by 5 I will get 1?

- 4 x 1 = 4 and dividing by 5 gives a remainder of 4, therefore since multiplying 4 by 1 does not give a remainder of 1, 1 cannot be the inverse of 4 (mod 5).

- 4 x 2 = 8 and dividing by 5 gives a remainder of 3, therefore since multiplying 4 by 2 does not give a remainder of 1, 2 cannot be the inverse of 4 (mod 5).

- 4 x 3 = 12 and mod 5 gives 2 =/= 1 therefore 3 is not the inverse of 4 (mod 5)

- 4 x 4 = 16 and mod 5 gives 1! So when I multiply 4 by 4 (16) and divide by 5, I get a remainder of 1. Therefore I can conclude that the inverse of 4 in F5 is 4.

The same applies to the second part, I cannot read it properly.

This is the basic way of finding inverses but there is a better way that doesn't involve trial and error in case you are dealing with larger numbers.

February 18 2015, 6:35 pm

**0**

Hi Boitumelo.

I'll explain the basics now and then when you understand that part, we can go to the harder stuff. Basically what he is talking about it called modular arithmetic. It has got to do with divisions and keeping reminders.

When they say "4 modulo 3" or "4 mod 3", they mean divide 4 by 3 and write what the remainder is, in this case the answer is 1. "5 mod 2 = 3" and "6 mod 10 = 6".

So in F5 (field of 5), the modulo or mod is 5. So we are looking for the inverse of 4 in mod 5. The simplest definition of inverse is:

"An inverse of x is a number such that when you multiply x with it, you will get the number 1, for example the inverse of 2 is 1/2 because when you multiply the 2 and 1/2 you will get 1."

That is just a basic definition. It becomes a little bit complicated, not too much lol, when it comes to modular arithmetic. The easiest way to find an inverse mod something is to ask yourself "What number should I multiply x with such that when I divide by the mod, I get 1". Let's do your example:

What number should I multiple 4 with so that when I divide by 5 I will get 1?

- 4 x 1 = 4 and dividing by 5 gives a remainder of 4, therefore since multiplying 4 by 1 does not give a remainder of 1, 1 cannot be the inverse of 4 (mod 5).

- 4 x 2 = 8 and dividing by 5 gives a remainder of 3, therefore since multiplying 4 by 2 does not give a remainder of 1, 2 cannot be the inverse of 4 (mod 5).

- 4 x 3 = 12 and mod 5 gives 2 =/= 1 therefore 3 is not the inverse of 4 (mod 5)

- 4 x 4 = 16 and mod 5 gives 1! So when I multiply 4 by 4 (16) and divide by 5, I get a remainder of 1. Therefore I can conclude that the inverse of 4 in F5 is 4.

The same applies to the second part, I cannot read it properly.

This is the basic way of finding inverses but there is a better way that doesn't involve trial and error in case you are dealing with larger numbers.

February 18 2015, 6:35 pm

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