I was thinking about this interesting property of positive integers, where if a, b, are positive integers, and a>b, then the remainder when a and b are divided by a-b seems to always the same. Is this true for all positive integers? I can't seem to find a simple intuitive proof for this. I'm also curious if this is an iff relationship, if two positive integers a, b have the same remainder when divided by x, is x always a-b? thanks!

Guest Jun 15, 2019

#1**+3 **

Given that a and b are positive integers where a > b , I do think it is always true that

a mod (a - b) = b mod (a - b)

I don't know the best way to explain it, but here is how I convinced myself. In the diagram,

w is the remainder when a is "filled up" with as many blocks of width a-b as possible.

x is the remainder when b is "filled up" with as many blocks of width a-b as possible.

In other words,

w = a mod (a - b)

x = b mod (a - b)

And we can see that...

w + (a - b) = x + (a - b)

w = x

Therefore,

a mod (a - b) = b mod (a - b)

But I do not think it is always true that if a mod x = b mod x then x = a - b

For example, 10 mod 3 = 4 mod 3 but 3 ≠ 10 - 4

hectictar Jun 15, 2019

#5**+1 **

Thanks Hectictar,

I have been playing around with modular arithmetic lately,

It will take me a while to get my head around this.

I am not asking for more explanation, (not yet anyway) it is just a compliment.

Melody
Jun 16, 2019

#7**+1 **

I see. Thanks for all the help! I figured it had something to do with modular arithmetic, but I myself barely know the basics.

Guest Jun 17, 2019

#8**+2 **

This pretty much sums up my knowledge of it: http://mathforum.org/library/drmath/view/55771.html

hectictar
Jun 17, 2019

#2**+1 **

two positife integers a and b such that a>b have the same remainder when divided by x if and only if x is a factor of a-b

Guest Jun 15, 2019