
#1
Nov2213, 07:33 AM

P: 5

hey everybody, once i saw a thread here (didn't want to revive it) about an equation that proves that 1=1, it was proved wrong ofc, but at the end, someone posted this:
" 1=(1)^1 =(1)^2*1/2 =[(1)^2]^1/2 =(1)^1/2 =√1 =1 " yet no one replied to it, can someone show me which is the "trippy" step here? the one that misuses an algebra rule? (e.g. a rule that can only be applied to positive numbers etc..." thank you. 



#2
Nov2213, 07:47 AM

Mentor
P: 4,499

We get this question a lot, there's actually a thread in the FAQ devoted to answering questions like it
http://www.physicsforums.com/showthread.php?t=637214 The main point is that whne you write [tex] \left( 1 \right)^{2/2} = \left( (1)^2 \right)^{1/2} [/tex] you have performed an operation which is not actually valid. Taking exponents of negative numbers is tricky and you have to be more careful than when you are working with exponents of positive numbers. In general [tex] x^{ab} =\left( x^{a} \right)^{b} [/tex] is something that can only be applied when x is a positive number. 



#3
Nov2213, 07:54 AM

P: 5

okay thanks alot! i appreciate it! ^^




#4
Nov2213, 08:03 AM

P: 651

i need a proof that 1 DOES NOT equal 1may i know why the indices rule is invalid for negative numbers? i tried for example, (2^6) and split them up to [2^(2*3)] = 4^3 and i still yielded 64. where does taking exponents of negative numbers breakdown? thanks! 



#5
Nov2213, 09:14 AM

Mentor
P: 21,067

If you want to raise 2 to the 6th power, you have to write it as (2)^6. 



#6
Nov2213, 10:30 AM

P: 651

anyway, is the even root fractional exponent the only case whereby this rule breaks down ? 



#7
Nov2213, 07:40 PM

Mentor
P: 21,067

If you have an expression such as (27)^{2/3}, you can write it either as [(27)^{2}]^{1/3} or as [(27)^{1/3}]^{2}, both of which are equal to 9. The first expression simplifies to (729)^{1/3} = 9, and the second expression simplifies to (3)^{2}, which is also 9. 



#9
Nov2313, 05:43 AM

P: 651





#10
Nov2313, 08:38 PM

P: 10

Start with: [itex]\sqrt{x}[/itex] Now to factor out a 1: [itex]=i\sqrt{x}[/itex] And to factor out another 1: [itex]=i*i\sqrt{x}[/itex] [itex]=\sqrt{x}[/itex] [itex]\Rightarrow \sqrt{x}=\sqrt{x}[/itex] :P The issue is that even functions are not 11, meaning they can map multiple inputs to the same output. Naturally, the inverse function would have to map backwards, but it would have to be split off to multiple values. That is why, for example, [itex]\sqrt{9}=\{3,3\}[/itex] 



#11
Nov2413, 06:46 AM

P: 754





#12
Nov2413, 07:51 AM

P: 269

First you add ##1## to both sides of the equation ##1=1## and get ##1+1=0##. Next you use the order axioms to show that ##1+1>1>0##, which is a contradiction and proves that ##1## can't equal ##1## (for real numbers ##a## and ##b##, the inequalities ##a>b## and ##a=b## can't both be true). 



#13
Nov2913, 02:05 PM

P: 1,227





#14
Nov2913, 02:18 PM

P: 10




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