Assignment 0 : Problem 3

[Aaron Albers]

Prove that 3 + 5(2)^.5 is not the square of a real number of the form a + b(2)^.5, where a and b are integers.

[Aaron Albers]

I am not sure I understand the question.... anyone else know what is being asked?

[Anne S]

if 3+(5^.3) is x^2, is x a real number?

On Sep 6, 5:21 pm, Aaron Albers <aaroncalb...@gmail.com> wrote:
> I am not sure I understand the question.... anyone else know what is being
> asked?
>

[Aaron Albers]

In this case what does "real" mean. Is it the real domain rather than imaginary or is it whole numbers? If you solve that equation for x, x is in the real domain but it is not a whole number.

[Anne S]

Reviewing the inside cover and page 7 -- I think "Real" means "not
complex". Since the given form has two term I think you have to FOIL
the term to square it.

On Sep 6, 5:37 pm, Aaron Albers <aaroncalb...@gmail.com> wrote:
> In this case what does "real" mean. Is it the real domain rather than
> imaginary or is it whole numbers? If you solve that equation for x, x is in
> the real domain but it is not a whole number.
>

[Aaron Albers]

3 + 5(2)^.5 = (a + b(2)^.5)^2 = (a + b(2)^.5)(a + b(2)^.5) = a^2 + 2ab(2)^.5 + 2b^2 = (a^2 + 2b^2) + 2ab(2)^.5

3 + 5(2)^.5 = (a^2 + 2b^2) + 2ab(2)^.5

Therefore

a^2 + 2b^2 = 3 -> a = (3 - 2b^2)^.5

2ab = 5 -> ab = 2.5 -> b(3 - 2b^2)^.5 = 2.5 -> b^2(3 - 2b^2) = 6.25 -> 3b^2 - 2b^4 - 6.25 = 0 (This equation does not have any real roots)