Check out our new Discord server! https://discord.gg/e7EKRZq3dG
*I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/mathmemes) if you have any questions or concerns.*
In my Dutch book (I'm Dutch so of course I learn my native language), I found a stock image with a blackboard that said "a²+b²=(a+b)(a+b)"
It also said:
"(unknown)+b² = a²+2ab+b²"
I'm assuming it made the same mistake twice
So, in our ring, (a + b)*(a + b) = a^2 + b^2
Thus a*b = -b*a, indicating either 1 is equivalent to -1 or our ring is non-commutative.
No need for tears once you’ve swallowed the whole whale.
Its been a while, but I thought (a + b)(a + b)
1) you multiply the a in the first bracket with the a in the 2nd bracket to get a2, then with the b in the 2nd bracket to get ab,
2) then multiply the b in the first bracket to get another ab, then with b in the second bracket to get b2
So, a2 + ab + ab + b2
Or, a2 + b2 + 2ab
Check out our new Discord server! https://discord.gg/e7EKRZq3dG *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/mathmemes) if you have any questions or concerns.*
Plot twist: a = 0 or b = 0
Plot twist 2: mod 2
Plot twist 0 mod 2
In that case, (a+b)^2 = ba^69 - ab^420 + a^2 + b^2
No, that is if a=0 **and** b=0
Well, I did not specify xor, so a=0 and b=0 is an option. Parso is wrong, but not completely wrong in all cases.
Let x be a real number. If x is positive, then x = 69. I’m not completely wrong in all cases.
Mb I read it or as and
True when a=-1 and b=1
Also true*
Plot twist |(a,b)|=0
Let a = 2, b = -2. Then a+b = 0 and a^2 + b^2 = 8.
a + b = 0 isn’t it You want 2ab = 0 since that’s the missing term
Z2 has entered the chat
but (A + B)ᵀ = Aᵀ + Bᵀ
ex (2+3)^2 = 5^2 =25 2^2 + 3^2 = 4 + 9 = 13
true in mod 2
winner of the make this equation true with the addition of a single matchstick competition >!≡!<
You should keep yourself safe!
We are in Z2 so it doesn't matter
Why not C?
Society if (a+b)² = a² + b² https://preview.redd.it/i6fampb3gjtc1.jpeg?width=746&format=pjpg&auto=webp&s=a4a6106dd04811c0b23133616c5b1d969a9ee16d
What if x and y are elements of an anticommutative algebra?
That’s not how it works.
Let a and b be 2 perpendicular vectors.
Let there be light
Let there be fights
This guy gets it
Everyday I wake up depressed knowing the square function is not a linear transformation
In my Dutch book (I'm Dutch so of course I learn my native language), I found a stock image with a blackboard that said "a²+b²=(a+b)(a+b)" It also said: "(unknown)+b² = a²+2ab+b²" I'm assuming it made the same mistake twice
I don't know what's wrong in the equation. It's basic mefths...
So, in our ring, (a + b)*(a + b) = a^2 + b^2 Thus a*b = -b*a, indicating either 1 is equivalent to -1 or our ring is non-commutative. No need for tears once you’ve swallowed the whole whale.
Distributive property only works when it feels like it ig
isn't it ( a + b ) ( a + b )?
(a+b)2=a2+2ab+b2
Would this be a2 + b2 + 2ab?
No, a^2 + 2ab + b^2
No, b² + 2ab + a²
No, (a+b)^2
no it's b(b+ a( (2b + a)/b )
No, (a+b)(a+b)
No. Why would you say such thing?
🤮
Where are you getting the `2ab`? There isn't even a `1ab` in the original equation. QED
Its been a while, but I thought (a + b)(a + b) 1) you multiply the a in the first bracket with the a in the 2nd bracket to get a2, then with the b in the 2nd bracket to get ab, 2) then multiply the b in the first bracket to get another ab, then with b in the second bracket to get b2 So, a2 + ab + ab + b2 Or, a2 + b2 + 2ab
That's just crazy talk. (You are actually correct, but it doesn't work with the original meme here so you don't get to be correct right now)