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Oh… Oh dear…
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 missing 2*abe
it’s true in ℤ/2ℤ
(a+b)^2=a^2 + b^2 + 2abBy the cancellation law
a^2 + b^2 + 2ab = a^2 + b^2only when
2ab = 0So either
a=0orb=0erm 13! ≠ 25
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)^3By 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.)
True when a and b are orthogonal.
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
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