# Prove the sqrt(8) is not a rational number

• Sep 17th 2009, 06:14 AM
Bruins1
Prove the sqrt(8) is not a rational number
This is the proof I used. Does anyone know if it is correct?

Assume that the sqrt(8) is a rational number.

sqrt(8) = n/m Assume no common factors between n and m

8 = n^2/m^2

8m^2 = n^2

So n^2 must be an even number since anything muliplied by 8 is even.
n must be even as well because even numbers squared are even, odd numbers squared are odd.

Represent n as 4k
n^2 = (4k)^2 --> 4^2k^2 --> 16k^2

8m^2 = 16k^2
m^2 = 2k^2
So, m^2 is even
So, m is even
Therefore m and n have a common factor, it disproves my assumption
• Sep 17th 2009, 08:29 AM
Hello Bruins1
Quote:

Originally Posted by Bruins1
This is the proof I used. Does anyone know if it is correct?

Assume that the sqrt(8) is a rational number.

sqrt(8) = n/m Assume no common factors between n and m

8 = n^2/m^2

8m^2 = n^2

So n^2 must be an even number since anything muliplied by 8 is even.
n must be even as well because even numbers squared are even, odd numbers squared are odd.

Represent n as 4k
n^2 = (4k)^2 --> 4^2k^2 --> 16k^2

8m^2 = 16k^2
m^2 = 2k^2
So, m^2 is even
So, m is even
Therefore m and n have a common factor, it disproves my assumption

Perfect!

• Sep 17th 2009, 08:30 AM
xalk
Quote:

Originally Posted by Bruins1
This is the proof I used. Does anyone know if it is correct?

Assume that the sqrt(8) is a rational number.

sqrt(8) = n/m Assume no common factors between n and m

8 = n^2/m^2

8m^2 = n^2

So n^2 must be an even number since anything muliplied by 8 is even.
n must be even as well because even numbers squared are even, odd numbers squared are odd.

Represent n as 4k
n^2 = (4k)^2 --> 4^2k^2 --> 16k^2

8m^2 = 16k^2
m^2 = 2k^2
So, m^2 is even
So, m is even
Therefore m and n have a common factor, it disproves my assumption

8m^2 is even because 8m^2 = 2(4m^2).The proof is correct
• May 13th 2016, 10:23 PM
lakshaytaneja
Re: Prove the sqrt(8) is not a rational number // wrong proof
Quote:

Originally Posted by Bruins1
This is the proof I used. Does anyone know if it is correct?

Assume that the sqrt(8) is a rational number.

sqrt(8) = n/m Assume no common factors between n and m

8 = n^2/m^2

8m^2 = n^2

So n^2 must be an even number since anything muliplied by 8 is even. //true n^2 will be even
n must be even as well because even numbers squared are even, odd numbers squared are odd. // but if n^2 is even its not correct to say that n will be too as the square root may or may not exist..its like saying 8 is even therefore sqrt(8) is even which is not true.

Represent n as 4k //even if n was even it will be represented as 2k not 4k like 6 is even but we cant write it as 4k
n^2 = (4k)^2 --> 4^2k^2 --> 16k^2

8m^2 = 16k^2
m^2 = 2k^2
So, m^2 is even
So, m is even
Therefore m and n have a common factor, it disproves my assumption

...
• May 14th 2016, 09:59 AM
Archie
Re: Prove the sqrt(8) is not a rational number
You are correct, the proof is false but also seven years old and so pointing it out is unlikely to be useful. If you write n=2a, cancel out the 2s and again point out out that k must be even if k^2 is even, finally writing k=2a, you arrive at the desired contradiction.