Oh... Oh dear...
this post was submitted on 16 Jan 2025
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NonCredibleDefense
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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.
I'm pretty sure that's part of the joke.
it’s true in ℤ/2ℤ
(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
erm 13! ≠ 25
True when a and b are orthogonal.
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.)
Self explanatory
All I know is Left Add Right Subtract.
But I do have a pretty rad cannon.
Oh no
(a-b)(a+b) = a^2 + b^2
(a+ib)(a-ib) = a^2 + b^2