A little stumped on a number theory question

you may try analyse this using prime factors decomposition of a and b respectively. you can put constrains on what [tex]\frac{b^n}{a^n}[/tex] looks like and hence deduce what must happen to b/a.

 

if you use [tex]\frac{b^n}{a^n}=\left(\frac{b}{a}\right)^n[/tex] then it would be quite obvious as for real numbers [tex]x^n = 1[/tex] means x =1 when n is not zero.

 

Way to complicated. Suppose there is a counter example. If there is one, then there is one with a as small as possible. If p is a prime dividing a, then p divides p^n divides b^n, hence p divides b, since p is a prime

But then a/p is a smaller counter example, which is a contradiction. Hence there are no counter examples.

 

if b isnt divisible by a, then b^n/a^n=(b/a)^n=(c/d)^n where gcd(c,d)=1
so we get that b^n isnt divisble by a^n.

 

I'm not clear on this. Are you saying that (b/a)^n= (c/d)^n for any mutually prime c and d? Surely not. And if not, what are c and d?

If b is NOT divisible by a then b= ak+ c for integer k and non-zero integer c.
Can you now show that bⁿ is not divisible by aⁿ? What is (ak+ c)ⁿ?

 

what im saying is because a doesnt divide b, then if we take the reduced fraction of b/a=c/d where (c,d)=1, then also b^n/a^n=(b/a)^n=(c/d)^n isn't an integer.
cause, (c^n,d^n)=(c,d)^n.

 

And have you proved raising a non-integer to a power means you get a non-integer? As far as I can see, that is assuming the answer.

 

isn't it implied from this (c^n,d^n)=(c,d)^n=1^n=1
so we have that (c/d)^n=c^n/d^n, where c^n and d^n have no common factors, thus this fraction is in Q but not in Z.
the only thing that might be bad here, is that if (c^n,d^n)=(c,d)^n is a consequence of what i need to prove bacuse if it is so, then obviously it doesnt work.

 

If you're going to claim that then there is no reason to do any of what you did: the HCF of a^n and b^n is a^n, thus impliying that HCF of (a,b) is a automatically.

 

yes i can see your point, is it a valid proof though?

 

btw, outside britain are there any others who use the term hcf, i thought everyone is using gcd. (-:

 

To avoid being circular you have to rely on a result which already implies the required result in far fewer steps. That doesn't make it invalid, just unnecessary.

 

use contradiction. i.e. suppose a/b is not an integer, and prove no power of it is an integer.

 

it's quite easy
prove by contradiction

i am very suprised that such an EASY question has stumbled so many of you people. you should be ashamed of yourselves

 

what do you think i tried to do on post number 7?

 

who has it 'stumbled'*? No one as far as I can tell apart from the OP.


* I think you mean stumped, not stumbled. I can't believe that stumped you, figuring out that EASY piece of English.

 

adityab88, yiou should be ashamed of yourself for trying to embarrass people just because they do not solve a silly math problem.

there is no shame in not understanding math, but there is shame in trying to humiliate people.