
#1
Apr2610, 02:56 PM

P: 17

I think that I have proof of 1 being 1 and I can't find any flaw in it.
Could you please take a look? 1=i² => (1)²=(i²)² => 1 = i^4 => take the square root both sides 1 = i² i² = 1 v i² = 1 Thus proving 1 = 1 



#2
Apr2610, 03:06 PM

HW Helper
P: 1,495

To confuse you a little more can you find the mistake: [itex]2=\sqrt{4}=\sqrt{(2)^2}=2[/itex].




#3
Apr2610, 03:07 PM

P: 17

Yea I know those too xD
Does that mean that it is correct? (but just not used since it's crazy) Edit: the √(2)² is not 2, but 2 btw :P You probably meant (2)² = √4 = 2² 



#4
Apr2610, 03:41 PM

HW Helper
P: 1,495

One equals minus one?
No I meant exactly what I wrote, the root cancelling the square. We can't do this because we have defined taking the square root of a real number to be a positive value. This is called the principal square root. For the complex numbers this principal root is defined as [itex]\sqrt{z}=\sqrt{z}e^{i \pi/2}[/itex]. In general for complex numbers it is not even true that [itex]\sqrt{zw}=\sqrt{z}\sqrt{w}[/itex].




#5
Apr2610, 03:49 PM

P: 59

[tex]a^{bc}=\left(a^b\right)^c[/tex] is not generally true. For example [tex]\left(\left(1\right)^2\right)^{\frac 1 2}\neq1[/tex]. You should be careful with this rule when the base is not a positive real number and the exponent is not an integer.




#6
Apr2610, 03:51 PM

P: 17

Could you dumb it down a little? (I'm a collage student) 



#7
Apr2610, 03:59 PM

HW Helper
P: 1,495

We can write every complex number z in the form [tex]z=ze^{i \theta}[/tex] with z the distance between z and the origin and [itex]\theta[/itex] the angle between the xaxis and z (polar coordinates). If you have had some complex numbers you should know this representation of a complex number. From this it follows that [itex]i=e^{i \pi/2}[/itex] and [itex]i^4=e^{2 \pi i}[/itex]. Now taking the square root of i^4 we get [tex]\sqrt{i^4}=e^{i \pi}=1[/tex].




#8
Apr2610, 04:03 PM

P: 367

Sqrt(x) is a function (input/output relationships are unique), so given a number (perhaps 9), Sqrt(9) will map to 3.. never 3. If Sqrt(9) could be either 3 OR 3, it wouldn't be a function. Even though (3)^2 = 9 = (3)^2, the root function is defined to take positive values and produce positive values. Edit: the √(2)² is not 2, but 2 btw :P You probably meant (2)² = √4 = 2² This is exactly what you kind of said.. sqrt( (2)^2 ) is indeed 2 since (2)^2 gives us 4, and by the definition of the function, we will get the positive possible "root" only. "You probably meant (2)² = √4 = 2²" You probably made some typing mistake here.. (2)^2 = sqrt(4) = 2^2?? 4 = 2 = 4? I don't know 


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