Modular Arithmetic Homework: 1/2*(x-4)(x-5) = 4(x-4)(x-5) (mod 7)

[Falken_47]

Homework Statement

Hi everyone, I have a problem in the following modular arithmetic operation

1/2*(x-4)(x-5) = 4(x-4)(x-5) (mod 7) ("=" means congruent in this expression)

Homework Equations

The Attempt at a Solution

I am completely lost on how the operation is valid. If the coefficient is an integer between 0 to 6 then I can easily find its inverse using euclid extended algorithm but the problem here is that the coefficient is 1/2, a rational number. Any help on this problem will be greatly appreciated!

 

[SteveL27]

Homework Statement

Hi everyone, I have a problem in the following modular arithmetic operation

1/2*(x-4)(x-5) = 4(x-4)(x-5) (mod 7) ("=" means congruent in this expression)

Homework Equations

The Attempt at a Solution

I am completely lost on how the operation is valid. If the coefficient is an integer between 0 to 6 then I can easily find its inverse using euclid extended algorithm but the problem here is that the coefficient is 1/2, a rational number. Any help on this problem will be greatly appreciated!

4 x 2 = 1 (mod 7)

So 1/2 = 4 (mod 7)

The integers mod 7 are a field so you can always divide by any nonzero number.

Another way to think of it is that 1/2 is 1 * 2^-1. What is the multiplicative inverse of 2? It's 4. So 1/2 = 4 (mod 7).

 

[HallsofIvy]

1/2= 4 (mod 7) because 4(2)= 8= 1 (mod 7) and that is, after all, the definition of "multiplicative inverse".

 

[Falken_47]

Oh alright i think I'm starting to get it. So if the question is 1/3*(x-4)(x-5) it is congruent to 5(x-4)(x-5) (mod 7) because 3 * 5 = 1 mod 7?

And what if the denominator is larger than the mod, say 1/30*(x-4)(x-5), should it be:
30^-1 * (x-4)(x-5)
and 30* 4 = 1 mod 7
therefore 1/30*(x-4)(x-5) = 4 * (x-4)(x-5) mod 7 ?