Proving Root n is Irrational: Perfect Square Affects Proof

In summary: But 12 is a root because it is the square of 2, which is a prime number. OK here is how i approached it:n can either be a prime number or a composite1) 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
  • #1
pankaz712
8
0
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.
 
Physics news on Phys.org
  • #2


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


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.

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.
 
  • #4


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.

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. 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 won't 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.
 
  • #5


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.

Yup, this is good. :) There's only one error, the word 'root' should read 'square root' instead.

2) if n is a composite. Let's 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?
 
  • #6


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)

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.
 
  • #7


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.

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.
 
  • #8


VietDao29 said:
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.
 
  • #9


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?

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'.
 

1. How do you prove that the square root of a number is irrational?

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.

2. What is the significance of the perfect square in this proof?

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.

3. Can this proof be applied to all numbers?

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.

4. Why is it important to prove that root n is irrational?

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.

5. Are there any other methods to prove the irrationality of root n?

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.

Similar threads

  • Precalculus Mathematics Homework Help
2
Replies
49
Views
3K
  • Precalculus Mathematics Homework Help
Replies
8
Views
916
  • Precalculus Mathematics Homework Help
Replies
1
Views
1K
  • Precalculus Mathematics Homework Help
Replies
21
Views
3K
  • Precalculus Mathematics Homework Help
Replies
11
Views
1K
  • Precalculus Mathematics Homework Help
Replies
12
Views
2K
  • Precalculus Mathematics Homework Help
Replies
14
Views
3K
Replies
31
Views
2K
  • Precalculus Mathematics Homework Help
Replies
14
Views
802
  • Precalculus Mathematics Homework Help
Replies
2
Views
2K
Back
Top