(a-b)(a+b) = a^2 + b^2
this post was submitted on 16 Jan 2025
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(a+ib)(a-ib) = a^2 + b^2
Just use the damn Pascal's triangle!
1 = 1 * 1 = (a+b)^0
1 1 = 1a + 1b = (a+b)^1
1 2 1 = 1a^2 + 2ab + 1b^2 = (a+b)^2
1 3 3 1 = 1a^3 + 3(a^2)b + 3a(b^2) + 1b^3 = (a+b)^3
By the way, I have submitted maths homework that had me continue it up to layer 25. The exercise didn't explicitly state we should use the binomial formula, so why work smart when you can work hard?
(To be fair I just used WolframAlpha to get the values. Had to write down all terms still and do some calculations.)
Oh... Oh dear...
All I know is Left Add Right Subtract.
But I do have a pretty rad cannon.
True when a and b are orthogonal.
That's... not true
2^2 = 4
3^2 = 9
(2^2 +3^2) = 13
(2 + 3)^2 = 5^2 = 25
13 != 25
Yeah, I guess I just don't like looking at his wildly cocked eyes.
It's ironic that I couldn't find a high-res version of this one.
it’s true in ℤ/2ℤ
I'm pretty sure that's part of the joke.
erm 13! ≠ 25
(a+b)^2=a^2 + b^2 + 2ab
By the cancellation law
a^2 + b^2 + 2ab = a^2 + b^2
only when
2ab = 0
So either a=0 or b=0
Self explanatory