A strange inconsistency with square roots

[ELB27]

Hi,
I have a question that came into my mind while solving some problems. If I have a constant times an expression in a square root like ##4\sqrt{16}## I can square the constant and push it into the square root: ##4\sqrt{16}=\sqrt{4^2 16} = 16##. But what if the constant outside of the square root is negative? Then I will get a contradiction: ##-4\sqrt{16} = \sqrt{(-4)^2 16} = \sqrt{4^2 16} = 16 ≠ -16##. Why is this happening? Clearly I cannot square negative numbers the same way as I do with positive ones, but why?
Any comments will be appreciated!

 

[jedishrfu]

You have to remember that when you take the square root of a positive number there are always two answers one positive and one negative.

 

[Mark44]

Hi,
I have a question that came into my mind while solving some problems. If I have a constant times an expression in a square root like ##4\sqrt{16}## I can square the constant and push it into the square root: ##4\sqrt{16}=\sqrt{4^2 16} = 16##. But what if the constant outside of the square root is negative? Then I will get a contradiction: ##-4\sqrt{16} = \sqrt{(-4)^2 16} = \sqrt{4^2 16} = 16 ≠ -16##. Why is this happening? Clearly I cannot square negative numbers the same way as I do with positive ones, but why?
Any comments will be appreciated!

For starters, let's limit the discussion to the square roots of nonnegative numbers.

A property of square roots that you're using is this: ##\sqrt{ab} = \sqrt{a}\sqrt{b}##. This is valid only when a and b are nonnegative.

The problem your work is that ##\sqrt{(-4)^2} \neq -4 ##, so the first equality in your equation above isn't valid.

 

[Mark44]

You have to remember that when you take the square root of a positive number there are always two answers one positive and one negative.

I disagree. When you take the square root of a positive number, you get a single (positive) number. For example, ##\sqrt{9} = 3##. It's not ##\pm 3##. The square root symbol represents the principal (i.e., positive) square root of its argument.

 

[jedishrfu]

I disagree. When you take the square root of a positive number, you get a single (positive) number. For example, ##\sqrt{9} = 3##. It's not ##\pm 3##. The square root symbol represents the principal (i.e., positive) square root of its argument.

You're right, I was thinking of square root of x^2 where we don't know whether x is positive or negative.

 

[Mark44]

You're right, I was thinking of square root of x^2 where we don't know whether x is positive or negative.

And it's the same here, meaning that you get one answer:

##\sqrt{x^2} = |x|##

If the value of x is not known, you can't say the following
##\sqrt{x^2} = x##

Also, the following isn't true
##\sqrt{x^2} = \pm x##

 

[jedishrfu]

And it's the same here, meaning that you get one answer:

##\sqrt{x^2} = |x|##

If the value of x is not known, you can't say the following
##\sqrt{x^2} = x##

Also, the following isn't true
##\sqrt{x^2} = \pm x##

Okay got it. I keep forgetting the principal square root is the convention.

http://mathworld.wolfram.com/PrincipalSquareRoot.html

This must be one of the reasons I majored in Physics and not Math.

 

[DivergentSpectrum]

if you define square root x^1/2 as the inverse function of square x², then yes, there will be two answers, and one will be the negative of the other.
However, if you remember your algebra teacher telling you about the "vertical line test"(if somewhere on the graph you have 2 outputs=y for one input=x), then square root is not a function at all, because it has 2 outputs for each input.In case your wondering, a justification for why a negative times a negative is a positive:

suppose the money in your bank account is increasing by x dollars per year.

then x*n=amount of money you made in n years.

if x was negative, then it would mean you are losing money
then x*n would be negative

Heres where this comes into play

If n=number of years was negative
and x=money per year was positve,
then x*n would be negative because it would show how much money you would lose if you went back in time(when you are actually gaining money as time moves forward).

so, if both x and n were negative?
then because x is negative, you are losing money,
but because n is negative, you are also going backward in time
therefore you are gaining money as you go backwards in time.

Its kinda like how if you play country music backwards the guy gets his truck fixed, gets married, goes sober, and his dog comes back to life. :p

 

[Mark44]

if you define square root x^1/2 as the inverse function of square x², then yes, there will be two answers, and one will be the negative of the other.

No, this isn't right, either. The squaring function is not one-to-one, so doesn't have an inverse function.

However, if you remember your algebra teacher telling you about the "vertical line test"(if somewhere on the graph you have 2 outputs=y for one input=x), then square root is not a function at all, because it has 2 outputs for each input.

 

[ELB27]

Thanks very much for the replies!
So if I understand correctly, I was unconsciously missing the step ##4\sqrt{16}=\sqrt{4^2}\sqrt{16}## even with the positive number and just focused on pushing it into the square root, and thus the flawed step with the negative case was ##-4\sqrt{16}≠\sqrt{(-4)^2}\sqrt{16}##?

 

[Mark44]

Thanks very much for the replies!
So if I understand correctly, I was unconsciously missing the step ##4\sqrt{16}=\sqrt{4^2}\sqrt{16}## even with the positive number and just focused on pushing it into the square root, and thus the flawed step with the negative case was ##-4\sqrt{16}≠\sqrt{(-4)^2}\sqrt{16}##?

Right.

We get questions about square roots like this fairly often, and there are a lot of misconceptions around this concept, that seem to lead to contradictions. One of these is this one:
##1 = \sqrt{1 * 1} = \sqrt{(-1)(-1)} = \sqrt{-1} \sqrt{-1} = i * i = -1 ##
which appears to show that 1 and -1 are the same number. This breaks down at the step ##\sqrt{(-1)(-1)} = \sqrt{-1} \sqrt{-1}## for reasons I already gave.

Another misconception that shows up a lot is that, for example, ##\sqrt{4} = \pm 2##. The root of this confusion, I believe, is that we're taught that every positive real number has two square roots. As already mentioned, though, by definition and common usage, the √ symbol means the principal, or positive square root.

 

[DivergentSpectrum]

im pretty sure if you put the numbers into polar coordinates
z=re^iθ
you don't need to be as careful with these steps.
so if
a=r_ae^iθ_a
and
b=r_be^iθ_b
and i want to find
sqrt(a)*sqrt(b)
then i have
(r_ae^iθ_a)^1/2(r_be^iθ_b)^1/2
then i can still use the property sqrt(a)sqrt(a)=sqrt(ab) to get:

(r_ar_be^iθ_ae^iθ_b)^1/2

because e^(x)e^(y)=e^(x+y) i can change this to:

((r_ar_b)e^i(θ_a+θ_b))^1/2
then because (e^x)^y=e^(xy), i can say:

(r_ar_b)^1/2e^{i(θ_a+θ_b)/2}now z is positive if θ=0, and negative if θ=pi
so, if both are negative, then (θ_a+θ_b)/2=pi
so we have e^ipi=-1
then because r_a is just |a| and r_b=|b|
we have -(|a||b|)^1/2, which is correct for sqrt(a)*sqrt(b) where a and b are negative, even though i used sqrt(a)sqrt(a)=sqrt(ab)

to be honest, its a lot more work than is needed :p

 

[pwsnafu]

then because (e^x)^y=e^(xy)

This is not true for arbitrary complex x or y. See wikipedia.

 

[DivergentSpectrum]

This is not true for arbitrary complex x or y. See wikipedia.

also i forgot to take the root of unity lol

i think the whole a^xb^x=(ab)^x thing can be reduced to the idea that if f(x)=a and b, it doesn't necessarily mean a has to equal b, but from what i read on the page it seems to be implying that the inconsistencies with
(e^x)^y=e^(xy)
can be explained and the real answer is

while the inconsistencies with
ln(b^x)=x*ln(b)

cant be explained with a multivalued interperetation. pretty strange

 

[1MileCrash]

In case your wondering, a justification for why a negative times a negative is a positive:

suppose the money in your bank account is increasing by x dollars per year.

then x*n=amount of money you made in n years.

if x was negative, then it would mean you are losing money
then x*n would be negative

Heres where this comes into play

If n=number of years was negative
and x=money per year was positve,
then x*n would be negative because it would show how much money you would lose if you went back in time(when you are actually gaining money as time moves forward).

so, if both x and n were negative?
then because x is negative, you are losing money,
but because n is negative, you are also going backward in time
therefore you are gaining money as you go backwards in time.

Its kinda like how if you play country music backwards the guy gets his truck fixed, gets married, goes sober, and his dog comes back to life. :p

That's a pretty interesting "intuitive" explanation, but the fact does follow immediately from the distributive property.

 

[nucl34rgg]

Square root always returns a non-negative number. It's also only defined on non-negative numbers. Not every step is reversible. When you square a number and then take its square root, you lose information about the sign of the number, since this will always return a non-negative number.

 

[statdad]

Mark44: Edited by adding quote tags for the statements that statdad is citing.

Square root always returns a non-negative number.

If you are referring to the principal square root (commonly written [itex] \sqrt x [/itex]), but even in the realm of real numbers something like 25 has two square roots: 5 and -5.

It's also only defined on non-negative numbers.

If you are working in the realm of reals for domain and range, yes.

Not every step is reversible. When you square a number and then take its square root, you lose information about the sign of the number, since this will always return a non-negative number.

Yes (again, in the reals), but this is more clearly seen as a feature of the fact that [itex] y = x^2 [/itex] is not a one-to-one function. (The others are also, in sense).

But the point of all this is?