Think we've got dividing and multiplying by zero wrong, lads.
It should be:
[math]a \times 0 = a[/math]
[math]a \div 0 = a[/math]
So, the exact rules for dividing and multiplying by 1.
Because:
[math]\sqrt{a^2 + b^2} = |a|\sqrt{(\frac{b}{a})^2 + 1}[/math]
Is true for everything, except 0 under our current rules for dividing and multiplying by zero, therefore either our current rules are wrong or it just doesn't work with 0.
Can Veeky Forums explain why I am retarded to think the former?
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Because your mother dropped you on your head when you were young
[math] 0
i clearly meant n^2 btw
>implying i^2 > 0
kys brainlet
it was implicit that i was talking about R
n usually implies an arbitrary element of the set of reals..
how did you get to know this?
n is more often a natural number or maybe an integer, not a real brainlet
i did my thesis on zero division
[math] a \times 0 = a \iff \\ a \times (a - a ) = a \iff \\ a^{-1} \times a \ (a-a) = a^{-1} \times a \iff \\ a-a = 1 \iff \\ 0 = 1 [/math]
Whoops, back to the drawing board.
Why? OP's example works with real and imaginary numbers.
If you followed the rules of PEMDAS, you would get 1 x 0 = 1, which using OP's logic gives 1 =1.
M: NxN => N
1. M(a,0) = 0
2. M(a,S(b)) = P(a,M,(a,b))
a * 0 = a
a * 1 = a
a * (1 + 0) = a
a * 1 + a * 0 = a
a + a = a
2a = 2
you fucked the distributivity and ton of other properties
So...
[math]
\frac{a}{a} = 1 \ or \ 0
[/math]
Doesn't make sense senpai.
did you actually check any example?
Your proof is incorrect. Faggot.
Just this one.
I'm sure it works for all others too.
[math]|a|\sqrt{(\frac{b}{a})^2 + 1}[/math]
how does that even work?