Can you tell me if I'm right? :)

My sister and I are currently arguing about something regarding square roots. We'd like someone here to tell use who's right and why? :)

Basically, the question is: Is is true that Square root of 16 = ±4?

My reasoning is pretty simple:

Since Square root of 16 is 16^1/2, using the opposite operation, we could prove that both 4 and -4 are equal to 16^1/2 since (4)² = 16 and (-4)² = 16.

My sister's reasoning (well it's not hers since she's too stubborn to write it down...) is pretty much this:

First of all, she brought that the square root of -16 doesn't exist. I don't see the point in that, but she did... What she's basically saying is that you are finding the square root of (±4)² which then means that in can only be 4 for some magic... Honestly, I doubt she even understands her own logic...

When answering, please take into account that I want an answer that is useful for maths a 16 years old would do. Please don't include Ph.D. level reasoning! :)

Thank you! :)

Re: Can you tell me if I'm right? :)

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Originally Posted by

**Snowmen** My sister and I are currently arguing about something regarding square roots. We'd like someone here to tell use who's right and why? :)

Basically, the question is: Is is true that Square root of 16 = ±4?

My reasoning is pretty simple:

Since Square root of 16 is 16^1/2, using the opposite operation, we could prove that both 4 and -4 are equal to 16^1/2 since (4)² = 16 and (-4)² = 16.

My sister's reasoning (well it's not hers since she's too stubborn to write it down...) is pretty much this:

First of all, she brought that the square root of -16 doesn't exist. I don't see the point in that, but she did... What she's basically saying is that you are finding the square root of (±4)² which then means that in can only be 4 for some magic... Honestly, I doubt she even understands her own logic...

When answering, please take into account that I want an answer that is useful for maths a 16 years old would do. Please don't include Ph.D. level reasoning! :)

Thank you! :)

By convention $\displaystyle \sqrt{16} = 4$. This convention holds for any use of numbers. But if we are talking about a variable $\displaystyle \sqrt{x^2} = \pm x$.

-Dan

Re: Can you tell me if I'm right? :)

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Originally Posted by

**topsquark** By convention $\displaystyle \sqrt{16} = 4$. This convention holds for any use of numbers. But if we are talking about a variable $\displaystyle \sqrt{x^2} = \pm x$.

-Dan

I agree with the convention thing but say you had a test that asked you whether square root of 16 = ±4 is true or false, what would you answer? We're not talking about a variable by the way. :)

Re: Can you tell me if I'm right? :)

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Originally Posted by

**topsquark** By convention $\displaystyle \sqrt{16} = 4$. This convention holds for any use of numbers. But if we are talking about a variable $\displaystyle \sqrt{x^2} = \pm x$.-Dan

Dan, that is simply wrong mathematically. $\displaystyle \sqrt{x^2} =|x|$ is correct.

Re: Can you tell me if I'm right? :)

The definition of "square root" as a function is commonly defined to be specifically the *POSITIVE* number that when squared gives the original number. Yes it is an extremely asymmetrical and imbalanced definition, but it is agreed upon so that "square root" can be a *function*. Functions gives only *one* value to each number.

So the square root of 16 is 4.

But if you define it as a multiple-valued function, it can be -4 as well. So it can depend on the context. Honestly the argument is simply a question of notation or definition, much like arguing whether the first positive number is 1 or 0. Its not really a mathematical question. On a different planet, it may very well be that the square root of 16 as a function is -4.

Quote:

I agree with the convention thing but say you had a test that asked you whether square root of 16 = ±4 is true or false, what would you answer? We're not talking about a variable by the way. :)

I would ask them to define "square root".

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Originally Posted by

**Plato** Dan, that is simply wrong mathematically. $\displaystyle \sqrt{x^2} =|x|$ is correct.

But the square root if -1 doesn't equal 1, the absolute value of *i*.

Re: Can you tell me if I'm right? :)

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Originally Posted by

**SworD** I would ask them to define "square root".

**That is not the question**.

The **square root of 4 ** is $\displaystyle \pm 2$.

**BUT** $\displaystyle \sqrt4=2$

Do not confuse the radical symbol with the words "square root". They are two different concepts.

Re: Can you tell me if I'm right? :)

Ah thanks for the correction. I don't know too much about the precise definitions either. :P In that case it is true that the square root of 16 is both 4 and -4, but 4 is the principal square root.

Re: Can you tell me if I'm right? :)

Quote:

Originally Posted by

**SworD** The definition of "square root" as a function is commonly defined to be specifically the *POSITIVE* number that when squared gives the original number. Yes it is an extremely asymmetrical and imbalanced definition, but it is agreed upon so that "square root" can be a *function*. Functions gives only *one* value to each number.

So the square root of 16 is 4.

But if you define it as a multiple-valued function, it can be -4 as well. So it can depend on the context. Honestly the argument is simply a question of notation or definition, much like arguing whether the first positive number is 1 or 0. Its not really a mathematical question. On a different planet, it may very well be that the square root of 16 as a function is -4

I would ask them to define "square root"

But the square root if -1 doesn't equal 1, the absolute value of *i*.

Any one who is up to date with current mathematical notation knows that $\displaystyle \sqrt{x}$ is used only if $\displaystyle x\in\mathbb{R}^+$.

So that $\displaystyle \sqrt{-1}$ **has no meaning.**

What has happen by way of *model theory*, we know that we can add the symbol $\displaystyle i$ such that $\displaystyle i^2=-1$ to the reals, thus we have the complex number set.

Re: Can you tell me if I'm right? :)

o.O That level of rigor is too much for me. But yeah, to the OP, you can ignore my post and just listen to Plato xD

Re: Can you tell me if I'm right? :)

Quote:

Originally Posted by

**Plato** **That is not the question**.

The **square root of 4 ** is $\displaystyle \pm 2$.

**BUT** $\displaystyle \sqrt4=2$

Do not confuse the radical symbol with the words "square root". They are two different concepts.

I should have remembered $\displaystyle \sqrt{x} = |x|$ But this one I didn't know. Thank you for both the correction and the concept here.

-Dan

Re: Can you tell me if I'm right? :)

Quote:

Originally Posted by

**SworD** The definition of "square root" as a function is commonly defined to be specifically the *POSITIVE* number that when squared gives the original number. Yes it is an extremely asymmetrical and imbalanced definition, but it is agreed upon so that "square root" can be a *function*. Functions gives only *one* value to each number.

So the square root of 16 is 4.

But if you define it as a multiple-valued function, it can be -4 as well. So it can depend on the context. Honestly the argument is simply a question of notation or definition, much like arguing whether the first positive number is 1 or 0. Its not really a mathematical question. On a different planet, it may very well be that the square root of 16 as a function is -4.

I would ask them to define "square root".

But the square root if -1 doesn't equal 1, the absolute value of *i*.

That wasn't what Plato said. he said $\displaystyle \sqrt{x^2}= |x|$. [itex]\sqrt{(-1)^2}= \sqrt{1}= 1[/itex].

Re: Can you tell me if I'm right? :)

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Originally Posted by

**HallsofIvy** That wasn't what Plato said. he said $\displaystyle \sqrt{x^2}= |x|$. [itex]\sqrt{(-1)^2}= \sqrt{1}= 1[/itex].

No you misunderstood, I wondered what would happen if you let x equal *i*, not -1. But Plato answered that anyway.