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How to prove that root n is irrational, if n is not a perfect square. Also, if n is a perfect square then how does it affect the proof.
Can you show us what you tried, so that we may know where you got stuck, and help you? It's against the forums' policies to provide help, unless you show us that you did put some efforts in solving the problem. So, just give the problem a try, and let's see how far you can get.How to prove that root n is irrational, if n is not a perfect square. Also, if n is a perfect square then how does it affect the proof.
OK here is how i approached it:Can you show us what you tried, so that we may know where you got stuck, and help you? It's against the forums' policies to provide help, unless you show us that you did put some efforts in solving the problem. So, just give the problem a try, and let's see how far you can get.
In case, you don't know where to start, then I'll give you a little push: try Proof by Contradiction.
Yup, this is good. :) There's only one error, the word 'root' should read 'square root' instead.OK here is how i approached it:
n can either be a prime number or a composite
1) if n is prime:
a) assume root n is rational. therefore root n = p/q( where p, q are co-prime)
then n= p^2/q^2,
p^2= n *q^2
n divides p^2, therefore n divides p. Now for some integer k, p=nk
(nk)^2 = n*q^2
nk^2= q^2
n divides q^2, therefore n divides q. Now n is a common factor for p & q. But we know that p & q are co-prime. Hence our assumption is wrong, root n is irrational.
Well, no, this is not true. Not every composite number can be split into the product of 2 primes. It can be a product of 3, 4 or even more primes. Say: 30 = 2.3.5 (a product of 3 primes)2) if n is a composite. Lets say it is product of two primes c1 & c2.
yes i see that not all composite numbers can be split into the products of only 2 primes. But the proof remains the same even if when the number can be split into " n" number of primes.Well, no, this is not true. Not every composite number can be split into the product of 2 primes. It can be a product of 3, 4 or even more primes. Say: 30 = 2.3.5 (a product of 3 primes)
Well, but what if the power of the prime is not 1? What I mean is, something like this:yes i see that not all composite numbers can be split into the products of only 2 primes. But the proof remains the same even if when the number can be split into " n" number of primes.
Yes...i believe u are right. I will surely check again and see. Please could u explain your method of proving it.Well, but what if the power of the prime is not 1? What I mean is, something like this:
200 = 2^{3}5^{2}.
I doubt that the proof still remains the same.
You can look at the post #5 above. Note that, what you are trying to show is that n must be a perfect square.Well, you don't really need to split it into 2 separated cases like that. Assume that: [tex]\sqrt{n}[/tex] is rational, i.e:
[tex]\sqrt{n} = \frac{a}{b} \mbox{ , where } a, b \mbox{ are co-prime}[/tex]
[tex]\Rightarrow n = \frac{a ^ 2}{b ^ 2}[/tex]
[tex]\Rightarrow a ^ 2 \vdots b ^ 2[/tex]
Now, let's think a little bit. From here, what can you say about a, and b?