How to prove that root n is irrational, if n is not a perfect square. Also, if n is aby pankaz712 Tags: root n is irrational 

#1
Feb611, 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.




#2
Feb611, 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. 



#3
Feb611, 04:17 AM

HW Helper
P: 1,422

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



#4
Feb911, 02:30 AM

P: 8

How to prove that root n is irrational, if n is not a perfect square. Also, if n is an 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 coprime) 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 coprime. 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 coprime) 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 coprime. Hence our assumption is wrong. 



#5
Feb911, 05:07 AM

HW Helper
P: 1,422

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 coprime}[/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? 



#6
Feb911, 06:22 AM

P: 8





#7
Feb911, 06:38 AM

HW Helper
P: 1,422

200 = 2^{3}5^{2}. I doubt that the proof still remains the same. 



#8
Feb911, 08:35 AM

P: 8




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