Discrete Math Question Involving Congruence Modulus

[Bashyboy]

Solve for x: 4x=6(mod 5)

Here is my solution:

From the definition of modulus, we can write the above as [itex]\frac{4x−6}{5}=k[/itex], where [itex]k[/itex] is the remainder resulting from [itex]4x~mod~5=6~mod~5=k.[/itex]

Solving for [itex]x[/itex], [itex]x=\frac{5k+6}{4}⟹x(k)=\frac{5k+6}{4}[/itex]

Now, my teacher said that is incorrect, and that [itex]k=...−2,−1,0,1,2,...[/itex]

I honestly don't understand what is wrong about my answer; and shouldn't k only take on nonnegative values, following from the definition of modulus?

 

[mfb]

and shouldn't k only take on nonnegative values, following from the definition of modulus?

k can be any integer number. 4=6 mod 2, and 6=4 mod 2.
"mod 2" refers to the "=", not to one side of the equation. This is different in computer science.

Your x(k) will give answers only if the fraction there (evaluated in the real numbers) gives integers.

 

[Bashyboy]

So, I presume there is a different way of solving this, of which I am unaware of? Could you possibly give me a hint?

 

[Bashyboy]

I think I might have discovered why I am unable to do this problem: my teacher furnished this as a test question, despite the fact that my teacher did not even assign the chapter which deals with solving congruences, nor did my teacher assign probems. Frustrating but not out of the ordinary.

 

[HallsofIvy]

Solve for x: 4x=6(mod 5)

Here is my solution:

From the definition of modulus, we can write the above as [itex]\frac{4x−6}{5}=k[/itex], where [itex]k[/itex] is the remainder resulting from [itex]4x~mod~5=6~mod~5=k.[/itex]

It would be better to write 4x- 6= 5k or 4x= 6+ 5k since all numbers here must be integers. And writing it as a fraction obscures that.

Now: if k= 0, this is 4x= 6 which does not have an integer solution.
If k= 1, this is 4x= 11 which does not have an integer solution.
If k= 2, this is 4x= 16 which has x= 4 as a solution.

And since this is "modulo 5", any x= 4+ 5k, for k any integer, is a solution.

Solving for [itex]x[/itex], [itex]x=\frac{5k+6}{4}⟹x(k)=\frac{5k+6}{4}[/itex]

Now, my teacher said that is incorrect, and that [itex]k=...−2,−1,0,1,2,...[/itex]

I honestly don't understand what is wrong about my answer; and shouldn't k only take on nonnegative values, following from the definition of modulus?

No, there is no requirement that "k" be non-negative. I don't know which of several equivalent defnitions of "modulus" you are using. One is "a= b (mod n) if and only if a- b is divisibe by n". Another is "a= b (mod n) if and only if a= b+ kn for some integer n".

4- 5= -1 is also a value of "x= 4 (mod 5)" and satisfies 4(-1)= -4= 6 (mod 5) because 6- (-4)= 10 is a multiple of 5.