Brennan.Tobias

hello

I did look at that @JulianRachman but I couldn't say anything about it

im not being rigorous just showing intuition

so its just x/x = 1

because near 0, sin(x) is like x

isn't that 1?

no it doesn't thats division by zero

no

if f and g are cont. at y then lim x->y (f(x)g(x)) = (lim x->y f(x))(lim x->y g(x))

@sharafzaman, you can do that at points a function is continous at

that will find you the other two 'extra' solutions

but you might also have 3|a-1 & 5|a+1 (or the reverse)

so clearly a=1 and a=-1 work

that's the same as 15 | (a-1)(a+1)

you want to solve a^2-1 = 0 mod 15

well another way to do it is this

@GGG is that any use to you

one easy way would be to try a=1,2,3,4,...,15

i see you're doing a different problem

sorry ignore that

could this be of any use to you math.stackexchange.com/questions/1661650/…

@ggg what are you working on?

this thread has become incredibly low quality math.stackexchange.com/questions/1665899/…

wow

thanks

I don't really want to argue about it

that the question was closed as duplicate

this was really a shame

hello

hi

ok thanks anyway

can someone help me with analytic number theory please?

I set a bounty on my prime gaps question

Hi

if so you could look over an answer to a question I asked?

is anyone interested in number theory/prime gaps

hi

Hello

hi

hello

hello I have a question about polynomials

bye!

if J is an ideal containing I

oh no it's just (R/I) / (J/I) = R/J

@ms.mop, im there's a theorem like (R/I)/(J/I) = R/(I+J) I think

hello

I'm just studying algebra

hey :)

hello