Question

Consider the following statement: 2 ^ (1/3) , the cube root of 2, is irrational.

(a) First, prove that if n^3 is even, then n is even, where n is an integer.

(b) Now, using a proof by contradiction, prove that 2^(1/3) , the cube root of 2, is irrational.

Answer 1

Ans: (a) We need to show that for all integers n, if n³ is even, then n is even.

We will proof this by contrapositive method.

So the contrapositive of this statement is For all integers n, if n is odd, then n³ is odd.

Let n be an odd integer.

Then n = 2k + 1 for some integer k.

It follows that,

n³ = (2k + 1)³

= (2k + 1)(2k + 1)²

= (2k + 1)(4k 2 + 4k + 1)

= 8k³ + 8k²+ 2k + 4k²+ 4k + 1

= 8k 3 + 12k 2 + 6k + 1

= 2(4k 3 + 6k 2 + 3k) + 1, which is odd.

Hence, this is true. So the statement if n³ is even, then n is even is true. Proved.

(b) Here we need to proof that 2^1/3 or cube root of 2 is irrational.

We'll proof this by contradiction method. So for this, let us take 2^1/3 is a rational number.

Then, 2^1/3 = p/q where p, q are integers and q$\neq$ 0 .

If HCF(p,q) $\neq$ 1 then by dividing p, q by HCF(p,q) , 2^1/3 can be recuded by

2^1/3 =a/b where HCF(a,b)=1

$\Rightarrow$ b.2^1/3 = a

$\Rightarrow$ b³.2 = a³

^{$\Rightarrow$}a³ is divisible by 2 i.e even

$\Rightarrow$a is divisible by 2. (from the above prf)

Let a = 2c , where c is an integer.

It means a³= 8c³

Hence, 2. b³ = 8c³

$\Rightarrow$ b³ = 4c³

This means b³ is even. And hence b is even.

$\Rightarrow$ a, b both are even.

But this is impossible because a/b is in simplest form which can not be reduced.

Hence the contradiction arises due to our wrong assumption.

So 2^1/3 is an irrational number. Hence prooved.