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How to prove that root n is irrational, if n is not a perfect square. Also, if n is a

by pankaz712
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pankaz712
#1
Feb6-11, 12:42 AM
P: 8
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.
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zketrouble
#2
Feb6-11, 01:44 AM
P: 47
If a root n is a perfect square such as 4, 9, 16, 25, etc., then it evaluating the square root quickly proves it is rational. The square root of the perfect square 25 is 5, which is clearly a rational number.

To prove a root is irrational, you must prove that it is inexpressible in terms of a fraction a/b, where a and b are whole numbers.

For the nth root of x to be rational:

nth root of x must equal (a^n)/(b^n), where a and b are integers and a/b is in lowest terms.


So the square root of 3 for example would be done like this:

2nd root of 3 must equal (a^2)/(b^2)

3 = (a^2)/(b^2)

3(b^2) = (a^2) So we know a^2 is divisible by three.
(b^2) = (a^2)/3 So we know b^2 is divisble by three.

Since this is neither of these are in lowest terms, the sqrt(3) is an irrational number.
VietDao29
#3
Feb6-11, 04:17 AM
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Quote Quote by pankaz712 View Post
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.

In case, you don't know where to start, then I'll give you a little push: try Proof by Contradiction.

pankaz712
#4
Feb9-11, 02:30 AM
P: 8
How to prove that root n is irrational, if n is not a perfect square. Also, if n is a

Quote Quote by VietDao29 View Post
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.
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. Lets say it is product of two primes c1 & c2. The proof remains the same for any number of primes.

a) assume root n is rational. therefore root n =p/q( where p,q are co-prime)
c1c2=p^2/q^2
p^2= c1c2*q^2
c1 divides p^2 therefore divides p. Now for some k, p= c1k
(c1k)^2 = c1c2*q^2
c1k^2 = c2* q^2
Now c1 divides L.H.S. , therefore must divide R.H.S. It wont divide c2( its a prime). Therefore it divides q^2. Therefore c1 divides q.
Now c1 is a common factor for p&q. But we know that p&q are co-prime. Hence our assumption is wrong.
VietDao29
#5
Feb9-11, 05:07 AM
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Quote Quote by pankaz712 View Post
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.
Yup, this is good. :) There's only one error, the word 'root' should read 'square root' instead.

2) if n is a composite. Lets say it is product of two primes c1 & c2.
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, 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?
pankaz712
#6
Feb9-11, 06:22 AM
P: 8
Quote Quote by VietDao29 View Post

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)
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
#7
Feb9-11, 06:38 AM
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Quote Quote by pankaz712 View Post
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, but what if the power of the prime is not 1? What I mean is, something like this:
200 = 2352.
I doubt that the proof still remains the same.
pankaz712
#8
Feb9-11, 08:35 AM
P: 8
Quote Quote by VietDao29 View Post
Well, but what if the power of the prime is not 1? What I mean is, something like this:
200 = 2352.
I doubt that the proof still remains the same.
Yes...i believe u are right. I will surely check again and see. Please could u explain your method of proving it.
VietDao29
#9
Feb9-11, 08:41 AM
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Quote Quote by VietDao29 View Post
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?
You can look at the post #5 above. Note that, what you are trying to show is that n must be a perfect square.

Now if [tex]a ^ 2 \vdots b ^ 2[/tex], what can you say about a, and b? In my proof, the symbol [tex]\vdots[/tex] stands for 'divisible by'.


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