Square roots

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  • #1
BruceW
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Main Question or Discussion Point

I understand the basics of raising numbers to powers of whatever, but i've been wondering:

When you square a number then square root it, you end up with plus or minus the original number. (i.e. 5 times 5 gets 25, then root gets 5 or -5). But using the rules of raising numbers to powers of other numbers:
(5^2)^0.5=5^1
And i would assume 5^1=just 5, not -5. But the left-hand-side of the equation gives 5 or -5.

Another example: 2^2=2^(4/2)=(2^4)^0.5=16^0.5= 4 or -4
But surely 2^2=4 and not -4

So there seems to be a contradiction. Is there some rule that defines when there is an uncertainy in the sign of the number?
 

Answers and Replies

  • #2
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[tex](2)^2=4[/tex] and [tex](-2)^2=(-2)(-2)=4[/tex]

[tex]\therefore[/tex] if [tex]x^2=4[/tex], x can be [tex]\pm 2[/tex]
 
  • #3
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When you square a number then square root it, you end up with plus or minus the original number.
No you don't: you end up with the modulus of the number:

[tex]\sqrt{x^2}=\left|x\right|[/tex]

By the way, these operations don't commute:

[tex]\left(\sqrt{x}\right)^2=x[/tex]

You are confusing the square roots of a number with the function [itex]f\left(x\right)=\sqrt{x}=x^{0.5}[/itex], so [itex]16^{0.5}=4[/itex], not 4 or -4.
 
  • #4
BruceW
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@JSuarez: So the function f(x) equals the positive root, so why don't the operations x and (x^0.5)^2 commute? They both have all the same answers for any given input. And (x^2)^0.5 would also commute, since we are saying the function equals the positive root only.
ps thanks for your help :)
 
  • #5
Mentallic
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@JSuarez: So the function f(x) equals the positive root, so why don't the operations x and (x^0.5)^2 commute? They both have all the same answers for any given input. And (x^2)^0.5 would also commute, since we are saying the function equals the positive root only.
ps thanks for your help :)
x and [tex](x^\frac{1}{2})^2[/tex] are the same. Take some positive x, you'll end up back at x after sqrt root then squaring. For some negative, you'll have a complex number then square it to get the negative back again.

However, when you square x first, you lose information as to whether x was positive or negative before squaring, so when you take the root it's not exactly x, but rather |x|.

When you square root x first, for x positive, you have some result positive (by the way, this is how square rooting has been defined, only as the positive square root, not plus/minus), but for x negative you have a complex number. You don't lose information in this case.

It's somewhat like taking the derivative of a function. Doing the inverse (integrating) means you lose the info as to what the constant was.
 
  • #6
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so why don't the operations x and (x^0.5)^2 commute?
When I said "commute", I meant that, given functions f(x) e g(x):

[tex]f\left(x\right)=\sqrt{x}[/tex]

[tex]g\left(x\right)=x^2[/tex]

Then [itex]f\circ g\left(x\right)=\left|x\right| \neq g\circ f\left(x\right)=x[/itex].
 
  • #7
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x and [tex](x^\frac{1}{2})^2[/tex] are the same.
No, they are not. As a counterexample, [itex]-2 \neq ((-2)^{1/2})^2[/itex]
 
  • #8
Mentallic
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No, they are not. As a counterexample, [itex]-2 \neq ((-2)^{1/2})^2[/itex]
[tex](-2)^{1/2}=\sqrt{2}i[/tex]

[tex](\sqrt{2}i)^2=-2[/tex]

Please explain why you're claiming they're not equivalent.
 
  • #9
HallsofIvy
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Mark44 would not have made such a mistake- it must be a typo.

I suspect he meant to say that [itex]((-2)^2)^{1/2}\ne -2[/itex].
 
  • #10
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It wasn't a typo. I was thinking in terms of the square root function defined on the nonnegative reals, but I didn't add this qualification. If you attempt to take the square root of a negative number on a calculator, the operation is undefined -- that's the sense of the square root I had in mind. Sorry I didn't make myself clearer.
 
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  • #11
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I understand the basics of raising numbers to powers of whatever, but i've been wondering:

When you square a number then square root it, you end up with plus or minus the original number. (i.e. 5 times 5 gets 25, then root gets 5 or -5). But using the rules of raising numbers to powers of other numbers:
(5^2)^0.5=5^1
And i would assume 5^1=just 5, not -5. But the left-hand-side of the equation gives 5 or -5.

Another example: 2^2=2^(4/2)=(2^4)^0.5=16^0.5= 4 or -4
But surely 2^2=4 and not -4

So there seems to be a contradiction. Is there some rule that defines when there is an uncertainy in the sign of the number?
Yes there's a way to solve this uncertainty, but using what's called "Complex numbers", where there's nothing called "not possible", like taking the square-root of a negative number.

If you would like to learn how this contradiction is solved, you have to learn the De Moivre's theorem, you can find details about it on wikipedia:

http://en.wikipedia.org/wiki/De_Moivre's_formula

In this theorem, every square-root gives you 2 outcomes, and every cube-root gives 3 outcomes, and every quad-rood gives 4 outcomes, and so on! So the number of outcomes depends on the power of the root you're applying. Those outcomes are not always nice and "symmetric" as they are for square-roots, positive and negative. They are sloppy somewhat and need graphical representation to make sense.

So the square-root of 1 is (1,-1). But the bad news is, that the cube-root of 1 is not only 1, but there are another two complex numbers, namely: cos(120) + i sin(120), and cos(240) + i sin(240). And "1", which is the first outcome we mentioned may be calculated with the same rule: cos(0) + i sin(0) = 1.

i is the kernel of the of the complex numbers, basically it's defined as the square-root of -1.

Sorry for complicating things, but this is why mathematics is nice, everything is symmetric, and everything has an answer, even if it doesn't make sense in the first sight :D

Hope this helps, any questions are welcome!

Ciao!
 
  • #12
Mentallic
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It wasn't a typo. I was thinking in terms of the square root function defined on the nonnegative reals, but I didn't add this qualification. If you attempt to take the square root of a negative number on a calculator, the operation is undefined -- that's the sense of the square root I had in mind. Sorry I didn't make myself clearer.
I also gave a mention to square roots of negatives, mainly:
x and [tex](x^\frac{1}{2})^2[/tex] are the same. Take some positive x, you'll end up back at x after sqrt root then squaring. For some negative, you'll have a complex number then square it to get the negative back again.
Calculators are not a nice way of disclaiming a fact.

For example, if you want to calculate the cotangent of a number that is equal to zero, most calculators don't have a cot function ready, so you need to take the reciprocal of the tangent function, but this leads to 1 divided by (undefined), but we know this is equal to 0:
[tex]\frac{1}{tan\left(\frac{\pi}{2}\right)}=0[/tex]

The calculator unsurprisingly gives us a math error since it tries to calculate [itex]tan(\pi/2)[/itex].

Similarly for the square root of a negative, just because the calculator gives us an undefined answer, doesn't mean it is undefined in the real sense.
 
  • #13
BruceW
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@ JSuarez:
Cool, that does make sense to me. So basically i just have to take care in the order in which i take those two functions.
Thanks, that's pretty much the answer i was looking for :)
 
  • #14
jgens
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Similarly for the square root of a negative, just because the calculator gives us an undefined answer, doesn't mean it is undefined in the real sense.
Mentallic, you're correct if you're working in the field of complex numbers, but if you're working strictly in the field of real numbers, then [itex](\sqrt{x})^2 \neq x[/itex] since negative reals have no square roots. I believe that this is Mark44's point when he is talking about calculators because most calculators work exclusively with real numbers.
 

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