MissKitten Posted October 29, 2010 Report Share Posted October 29, 2010 (edited) 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 October 29, 2010 by MissKitten Quote Link to comment Share on other sites More sharing options...
0 EDM Posted October 29, 2010 Report Share Posted October 29, 2010 It is sooooo true!!! Quote Link to comment Share on other sites More sharing options...
0 MissKitten Posted October 30, 2010 Author Report Share Posted October 30, 2010 yeah, i know. as soon as i can remember how to make 2x2=5, ill post that too. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 19, 2010 Report Share Posted December 19, 2010 [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)( = 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. Quote Link to comment Share on other sites More sharing options...
0 MissKitten Posted December 28, 2010 Author Report Share Posted December 28, 2010 Party pooper. jk! Quote Link to comment Share on other sites More sharing options...
Question
MissKitten
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 MissKittenLink to comment
Share on other sites
4 answers to this question
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.