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MissKitten

Question

Thought this was cool, but i didnt know where to post it, so here it is!

[sqrt(x)]=the square root of x

Given:

[sqrt(a)(b)]=[sqrt(a)][sqrt(b)]

[sqrt(-1)]=i

1+1

=1+[sqrt(1)]

=1+[sqrt(-1)(-1)]

=1+[sqrt(-1)][sqrt(-1)]

=1+(i)(i)

=1+i2

=1+(-1)

=0

yeah, i know there is a small flaw, but i still think its pretty cool!

Edited by MissKitten

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yeah, i know. as soon as i can remember how to make 2x2=5, ill post that too.

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[sqrt(a)(b)]=[sqrt(a)][sqrt(b)]

It is pretty cool. I wonder what percentage of high school students know that the rule that sqrt((a)(B)) = sqrt(a)sqrt(b) only applies when a and b are non-negative real numbers (that's the hidden mistake for those who weren't sure). I myself hadn't considered it. I only realized it when following the work to show that 1+1=0.

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