Modular arithmetic 2

[mbbx9]

i think im subscribing too!!

i wish my lecturer would go through this stuff, rather than assuming it is obvious

[4thKyuubi]

I think.....

Im subscribing :)

[ShivMan007]

loveeee it!

[brandojazz]

I was wondering, for the addition property, I think it is quite clear that x=y(mod n) is the same as x+a=y+a(mod n), i.e. the way I view it is that, they yield the same reminder when the difference is divided by n. I was wondering, for the second one, ax=ay(mod n), I know that n|x-y and that n|a(x-y), but does it yield the same remainder? and if it does, can u prove it does? my intuition tells me it does yield the same remainder, but Im not sure yet...

[TheMathsters]

@brandojazz No: x-y and a(x-y) do not necessarily yield the same remainder when dividing by n. If x-y yield remainder r then a(x-y) will yield remainder ar. However, we only need to know that if x-y yield remainder 0 then a(x-y) also yield remainder 0.

[valdas0]

I have never seen a teacher as energetic as you.

[themathematician1]

Its ok to say 5|x-y or 5|y-x since if a is congruent to b mod m then b is also congruent to a mod m.

[Nathanielguy]

Your lectures are great! I would love if I'd had more teachers like you in school.

[loverofbeats]

thank you!!! :D

[Mr81points]

i love you.

[mathswithmel]

Thank you so much for this video :)

[yejiahua]

can i say "5|x-y" instead of "5|y-x"? i know they are the same but which one is the convention?

[02101987abhishek]

Yeah I would want to know this. Can't it be x -y ?

[TheMathsters]

@yejiahua Yes it's fine to say either of those as they are equivalent. I was always taught second one as convention which is why I use it, but I can't think of a good reason for it.

[andaninuman]

thank you very much! without this i would have failed my math test

[rolltiderobert]

this is great! very helpful