# Question about mod function and congruences

1. Apr 23, 2013

### John112

Im having a bit of hard time understanding how is that two intergers (a and b) divided by a common divisor (m) have the same remainder imply that the difference of (a and b) will aslo be divisible by m?

a $\equiv$ b (mod m) $\Rightarrow$ m|(a-b)

the "|" means divides just incase you aren't fimiliar with that symbol.

a $\equiv$ b (mod m) says a/m and b/m will have the same remainder. Since, they have the same remainder (a - b) will also be be divisible by m.

example 1) 29 $\equiv$ 15 (mod 7) $\Rightarrow$ 7|(29 -15)

Why is the difference of 29 -15 also divisible by 7?

Is it because when since 29 and 15 have the same reminder means that we are simply taking out factors of 7 and the common reminder from the 29 and 15?

29 - 15
[7(4) + 1] - [ 7(2) + 1]

= 7(2) = 14 which is divisible by 7

example 2) 11 $\equiv$ 4 (mod 7) $\Rightarrow$ 7|(11 - 4)
11 - 4
[7(1) + 4] - [7(0) + 4]

= 7 which is divisible by 7

Even if my reasoning is correct, please try to explain in your own way. I can do it mathmetically but that problem im having is understanding it.

Last edited: Apr 23, 2013
2. Apr 23, 2013

### tiny-tim

Hi John112!

If a and b are both = 4 (mod 7),

then there exist integers p and q such a = 7p + 4, b = 7q + 4,

so a - b = 7(p - q).

3. Apr 23, 2013

### John112

Thanks for that clear explanation tiny-tim!