Re: Prove by Contradiction

if a, b, c are consecutive integers, then if a is an even number, b is an odd number, and c is an even number.

$\displaystyle a = 2k \text{ } b = 2k + 1 \text{ } c = 2k + 2 $

(even number)^3 = even number

$\displaystyle a^2 = (2k)^3 = 8k^3 = 2(4k^3) $

(odd number)^3 = odd number

[tex] (2k+1)^3 = (4k^2 + 4k + 1)(2k+1) = 8k^3 + 4k^2 + 8k^2 + 4k + 2k + 1 = 8k^3 + 12k^2 + 6k + 1 = 2(4k^3 +6k^2 + 3k) + 1

even number + odd number = odd number

2k + 2m + 1 = 2(k+m) + 1

but if c is an even number, (even)^3 = even, so then this cant hold.

if u start with an odd, even, odd

then just calculate,

$\displaystyle (2k+1)^3 + (2k+2)^3 = 16k^3 + 36k^2 + 30k + 9$

$\displaystyle (2k+3)^3 = 8k^3 + 36k^2 + 54k + 27$

So $\displaystyle (2k+1)^3 + (2k+2)^3 - (2k+3)^3 = 8k^3 - 24k - 18 = 0$

So if you calculate the discriminat, this eqn. has only 1 real root. Then you can use another test to see the one real root isn't an integer (remember k must be an integer)

The calculation is tedious, use wolfram alpha, just type in (2k+1)^3 and all is done for you.

Re: Prove by Contradiction

That's great, thanks

+rep

Re: Prove by Contradiction

Could you also help me with these questions?:

Prove that there exists a unique real number x such that ln x = 2

&

Let a and b be non zero integers. Prove that a divides b and b divides a if and only if a = +/- b