Not homework

[Anonymous No.4531821]

Prove that there isn't a rational number with the square of 2

[Anonymous No.4531825]

assume sqrt2=p/q (p,q)=1

[Anonymous No.4531845]

let x be an element of the real numbers
and assume
<span class="math">
x^2 = 2
[/spoiler]

Then x = a/b where a and b are integers relatively prime to each other. a and b are not both even as they would have the common factor of 2. a and b MUST both be even and by showing this contradiction we will prove there is no rational number with the square of 2.

Since
<span class="math">
x^2 = 2
[/spoiler]

it follows that
<span class="math">
a^2/b^2 = 2
[/spoiler]
thus
<span class="math">
a^2 = 2b^2
[/spoiler]

Actually I'm too tired to finish and my jsmath skills probably suck

[Anonymous No.4531868]

>>4531845
mind summarizing where you were going with this?

[Anonymous No.4531905]

>>4531868
2b^2 is even, so a^2 is even, and a is even because if a were odd, a^2 = a(a) would be odd. This means a is an integral multiple of 2, say a = 2y, where y is an integer. Then 2b^2 = a^2 = 4y^2, then divide by 2 and b^2 = 2y^2. Bjut 2y^2 is even, thus b^2 is even, and b is even. Hence a and b are both even.

[Anonymous No.4531952]

>>4531905
quite clever

i think

i cant begin to follow that without my brain hurting

[Anonymous No.4531960]

x = a/b
x^2 = 2

a^2 = 2b^2
With a and b integer
This means that if I have a square with b*b apples, if I had b^2 more apples, I can get a square of size integer a.
The number of added apples will be b^2
On the top of the b square I add (a-b)*b = ab-b^2 apples. Integer.
On the left I add ab-b^2 apples too. Positive number since a>b. Makes sense for now.
I am left with b^2-2(ab-b^2) apples to fill the new top-left corner.

b^2-2ab+2b^2
= 3b^2-2ab < b^2
And a = b*sqrt(2)
So 3b^2- 2ab = sqrt(3)*b^2 - 2*sqrt(2)*b
Solving the polynom to know how many apples I have left there
sqrt(3)*b^2 - 2*sqrt(2)*b
delta = 8 > 0 so 2 solutions
b = (2*sqrt(2) +/- sqrt(8))/2*sqrt(3)
But sqrt(8) = sqrt(4*2) = 2*sqrt(2)
So b = 0. Can't be that because I need apples.
Or
b = (4*sqrt(2))/(2*sqrt(3))
= 2*sqrt(2)/sqrt(3) = 2*sqrt(2/3)
Can't work because sqrt(2/3) isn't an integer (derivative of sqrt is positive and sqrt(1) = 1..)
So we can't have an integer number of apples to add in the corner
so we can't have a perfect square of nice apples
so we can't have a^2 = 2b^2 with a,b integers
so we can't have x = a/b with x^2 = 2 and a/b integers

[Anonymous No.4531974]

Assume <span class="math">\sqrt{2}=\frac{p}{q}[/spoiler] with <span class="math">p,q[/spoiler] coprime integers. Then we get <div class="math">2=\frac{p^2}{q^2}\Rightarrow p^2=2q^2\Rightarrow 2|p^2\Rightarrow 2|p\Rightarrow 4|p^2\Rightarrow 2|q^2\Rightarrow 2|q</div> which is a contradiction to the assumption that <span class="math">p,q[/spoiler] are coprime. QED

[Anonymous No.4531987]

OP, when the pythagorean school discovered this, they murdered the person who made the discovery.