Prove the cube of any odd positive integer is also odd while the cube of any even positive integer is even too

Question

Then prove the cubed root of 2 is irrational using the previous answer and/or solution.

Answer 1

If you know about unique factorization into primes, the proof is pretty obvious. Even if you don't, the cube of an even integer 2k is 8k^3, which is obviously divisible by 2. And the cube of an odd integer 2k+1 is 8k^3 + 12k^2 + 6k + 1, which is odd (not divisible by 2), since the first three terms ARE divisible by 2 but the fourth one is not.

Now, suppose the cube root of 2 were rational. Then it could either be written as a fraction a/b where a is even and b odd, or else c/d where c is odd and d even.

In the second case, c^3 = 2d^3, but one side of that is odd and one is even, so we have a contradiction.

In the first case, a^3 = 2b^3. Suppose a = 2j (a is even, so there is such a j). Then b^3 = 4j^3. But b and hence b^3 are odd. That's also a contradiction.

Q. E. D.

Answer 2

Prove by Induction :
Base case : (1 ^3) = 1
(2 ^ 3) = 8

Let the statement be true uptil k = n
To prove the same for k = n + 1

case 1) if n is even : given n^3 is even
n + 1 is odd .
Its cube is (n+1) ^ 3 = n^3 + 1 + (3 * n * (n+1) )
(3 * n * (n + 1) ) is even
so (n+1)^3 = even + 1 + even = odd ;

Similarly can the other case be solved

Actually Induction is a strong way to prove many things , Given you know some base cases .

Answer 3

Odd cubed;
odd times odd = ODD, ODD times odd =O.D.D

Even cubed;
even times even= EVEN, EVEN times even= E.V.E.N

((can't b bothered to do the other bit lol (Excuse the capitals, just representing the different answers lol))

Answer 4

(2n+1)^3= 8n^3 +12n^2+6n+1 The three first summands are even so +1 the sum is odd
(2n)^3 = 8n^3 = 2( 4n^3) is even