- #1
John112
- 19
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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?
Essentially what I am asking is:
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 7example 2) 11 \equiv 4 (mod 7) \Rightarrow 7|(11 - 4)
11 - 4
[7(1) + 4] - [7(0) + 4]
= 7 which is divisible by 7Even if my reasoning is correct, please try to explain in your own way. I can do it mathmetically but that problem I am having is understanding it.
Essentially what I am asking is:
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 7example 2) 11 \equiv 4 (mod 7) \Rightarrow 7|(11 - 4)
11 - 4
[7(1) + 4] - [7(0) + 4]
= 7 which is divisible by 7Even if my reasoning is correct, please try to explain in your own way. I can do it mathmetically but that problem I am having is understanding it.
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