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is √(2/3) irrational?

 Jul 1, 2016
 #1
avatar+206 
+1

Irrational numbers are numbers that cannot be written as a fraction and it cannot be a repeating (numbers that repeat in the same order, For example .12121212) or terminating decimal (numbers that repeat, in the same order, but not in the entire number. For example .121234567). The square root of 2/3 = 0.816496580927726. This number does not have any numbers repeating so it is irrational. But I haven't done these kind of things in forever, so you might want to get someone to confirm it.

 Jul 1, 2016
 #2
avatar+33616 
+5

Suppose it were rational, then you could write it as the ratio n/m where n and m are relatively prime integers (ie  they don't have any factors in common), and m>n.Then by squaring both sides you would get

 

n^2/m^2 = 2/3.  So 3n^2 = 2m^2

 

The right hand side must be an even integer (because multiplied by 2), so therefore the left hand side must be as well. Since 3 is odd, n^2 must be even, which means n must be even (the square of an odd number is odd).  If n is even we can write n = 2p, where p is another integer. Hence we would have. 3*4p^2 = 2m^2. or

 

2*3p^2 = m^2

 

Now the left hand side is clearly even, so therefore m^2 must also be even, hence m must be even. 

 

But if m and n are both even, they can both be divided by 2. This contradicts our original assumption that n and m are relatively prime. We cannot have a contradiction, hence our original assumption that sqrt(2/3) is rational must be wrong.

 

sqrt(2/3) is irrational

.

 Jul 1, 2016

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