/sci/ - Science & Math

[Anonymous No.10157804]

>(a+b)2 [math]\neq[/math] a2+b2
what is this bullshit?

[Anonymous No.10157808]

>>10157804
[math]a^2 + b^2 \neq (a + b)^2[/math]
>4chan reformats superscript unicode

[Anonymous No.10157816]

>>10157808
[math]>he doesn't just use one math tag to say everything[/math]

[Anonymous No.10157834]

>>10157804
>not working in char 2

[Anonymous No.10157835]

Put numbers in and try it dickhead.

[Anonymous No.10157842]

>>10157804
>this board is blue
>still on 4chan domain
>not on 4channel
what is this bullshit

[Anonymous No.10157857]

>>10157835
a=b=0
Who is the dickhead now?

[Anonymous No.10157860]

Math would be a lot nicer if it did work

[Anonymous No.10157887]

>>10157804
Think of it geometrically

[Anonymous No.10157894]

>>10157804
this works tho..

[Anonymous No.10157900]

>>10157887
What does the "triple equal symbol" mean if eight is equal to the letter D?

[Anonymous No.10157929]

>>10157900
It's an equivalence class. A lot of times shorthand for congruence or any other equivalence class being referenced in the discussion.

[Anonymous No.10158682]

[math]7^2 + 10^2 \neq 17^2[/math]

[Anonymous No.10158699]

>>10157857
Try for cases a, b > 0. This is stated for all integers

[Anonymous No.10158720]

>>10157894
Do know basic algebra?
(a+b)^2 = a^2+2ab+b^2

[Anonymous No.10158727]

[math] \displaystyle
\color{red}{\mathsf{\text{(USER WAS BANNED FOR THIS POST)}}}
[/math]

[Anonymous No.10158741]

>>10158720
>+2ab

[Anonymous No.10158753]

>>10157804
it does hold in mod 2

[Anonymous No.10158987]

>>10157804

(3 + 4)2 != 3(2) + 4(2)
14 != 14

Think you're being dumb OP.

[Anonymous No.10159079]

>>10157804

[math] \text{Let } (F, +, *) \text{ be a field.} [/math]
[math] \text{Let } \phi: F \to F \text{ be a mapping} \\ \text{If }+\text{ has the morphism property under }\phi\text{ then }\forall a, b\in R: \phi(a+b) = \phi(a) + \phi(b).[/math]
[math] \text{However, if } \forall a \in R: \phi(a) = a^2, \text{then } \phi(a+b) = (a+b)^2 = (a+b)(a+b) = a(a+b) + b(a+b) = a^2 + ab + ba + b^2 = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2 = a^2 + 2ab + b^2 = \phi(a) + 2ab + \phi(b) [/math]
[math] \text{If }e_1 \text{ and }e_2 \text{ are both identity elements, then by definition of the identity element: }e_1 = e_2+e_1 = e_2 \text{ thus } e_1 = e_2 [/math]
[math] \text{Since } 0 \text{ is and additive identity as in } x + 0 = x \text{ that means that }0 \text{ is the unique additive identity.} [/math]
[math] \text{Therefore, if } 2ab \neq 0 \text{ then, }a^2 + 2ab + b^2 \neq a^2 + b^2, \text{because if they were equal then there would exist another additive identity other than }0. [/math]
[math] \text{Now since } 0*x \text{ then }0*x = (x-x)*x = x*x - x*x = 0 \text{ it follows that any number multiplied by } 0 \text{ equals }0[/math]
[math] \text{And because the inverse of } 2 \text{ is }2^{-1}=\frac{1}{2}, \text{ then if }2a=0, \text{ it follows that }(2a)*2^{-1} = (2*a)*2^{-1} = 2*(a*2^{-1}) = 2*(2^{-1}*a) = (2*2^{-1})*a = 1*a = a; \text{ and } 0*2^{-1} = 0, \text{ then if one multiplies both sides by }2^{-1}, \text{ we get that }ab = 0[/math]
[math] \text{Therefore }a^{-1}(ab) = a^{-1}0 = 0 \text{ and }a^{-1}(ab) = (a^{-1}a)b = 1b = b. \text{ So if the inverse }a^{-1} \text{ exists, then }a\neq 0 \text{ and } b=0, \text{ and since a field is left and right associative, it follows that if }ab=0, \text{ then }a=0 \text{ or }b=0[/math]
[math] \text{Because }a*a \in F, \text{ then if F is a non-trivial division ring, then }\neg \forall x,y \in F: x*y = 0 \text{ that is } \exists x,y \in R: \neg (x*y = 0)[/math]
[math] \text{Therefore }a^2 + 2ab + b^2 \neq a^2 + b^2 [/math]

[Anonymous No.10159526]

>>10159079
>Assuming 2≠0

[Anonymous No.10159535]

>>10157804
>>10157808
>>10158699
Doesnt say anything about any specific field or set does it?

Good job moving the goalpost

[Anonymous No.10160286]

>>10159526
Oh lol I didn't realise :/

[Anonymous No.10160312]

>>10157804
Fucking based op. I agree. Why the fuck is this bullshit?

[Anonymous No.10160314]

>>10158682
Prove it muhfukker

[Anonymous No.10160318]

>>10159079
>wasting this much time trying to prove something on a shitpost

[Anonymous No.10160320]

hi, what is this thread?

[Anonymous No.10160321]

>>10159526
Though if x = ab, then if 2ab = 0 it means that ab + ab = 0, but this is only true if ab is equal to the additive identity, and since there is only one additive identity ab = 0 :}

[Anonymous No.10160337]

>>10160320
It's about (a+b)^2 is surprisingly not being equal to a^2+b^2

[Anonymous No.10160346]

good thread

[Anonymous No.10160682]

(2+2)^2 is?
2^2 + 2^2 is?

[Anonymous No.10160809]

>>10157804