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pankaz712
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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.
pankaz712 said: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.
VietDao29 said: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.
pankaz712 said: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.
2) if n is a composite. Let's say it is product of two primes c1 & c2.
VietDao29 said: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)
pankaz712 said: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.
VietDao29 said: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.
VietDao29 said: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?
To prove that the square root of a number is irrational, we use a proof by contradiction. This means we assume the opposite of what we want to prove, and then show that it leads to a contradiction. In this case, we assume that the square root of a number, let's say n, is rational i.e. it can be expressed as a ratio of two integers. Then, we use algebraic manipulations to show that this assumption leads to a contradiction, thus proving that the square root of n is irrational.
The perfect square plays a crucial role in this proof because it helps us show that the square root of a number is irrational. We use the fact that perfect squares have an even number of factors to derive a contradiction, which ultimately proves that the square root of a non-perfect square number is irrational.
Yes, this proof can be applied to all numbers, as long as we are trying to prove that the square root of a non-perfect square number is irrational. This proof cannot be applied to perfect square numbers, as their square roots are rational.
Proving that root n is irrational helps us understand the nature of numbers and their properties. It also has practical applications in fields such as cryptography, where the security of certain algorithms depends on the irrationality of certain numbers.
Yes, there are other methods to prove the irrationality of root n, such as the continued fraction method or the Euclidean algorithm method. However, the proof by contradiction method used in this particular proof is one of the most commonly used methods and is applicable to a wide range of numbers.