Jump to content
Sign in to follow this  
Muon321

Question about square roots!

Recommended Posts

Is it mathematically correct to say √16=±4, or would it be correct to say that √16=4, unless specified ±√16? This isn't for any specific problem or anything, but I'm just curious if is correct to have ±, because I know every number has two square roots.

Share this post


Link to post
Share on other sites

No it would not be correct to say that the square root of 16 is plus / minus 4.

 

That is simply because you are using slack wording. "the square root" is singular whereas plus / minus 4 is plural.

 

Yes the number 16 has two square roots +4 and -4.

 

What you are thinking about is the square root function y= sqrt(x). Functions are defined to have only one value so we take the positive ones by default and define a second function z=-sqrt(x) to access the engative values.

 

16, of course is a number, not a function.

Share this post


Link to post
Share on other sites

You're half right studiot. When you are doing basic algebra and solving via "undoing" method, you actually use plus or minus 4. BUT, when you are dealing with coordinates on a function or coefficients of a function you you described, you are correct, because using both roots would not create a function in a Cartesian plane, vertical line test fail.

Share this post


Link to post
Share on other sites

In case there's any confusion, the answer to OP's question is that [math]\sqrt{16}=4[/math]. The notation [math]\sqrt{x}[/math], when x is positive, refers to the principal square root, which is also positive. If you want to refer to both square roots of 16, then you should write [math]\pm \sqrt{16}[/math]. As studiot mentioned, "the square root" usually refers to the principal square root, also.

Share this post


Link to post
Share on other sites

.

 

Edited by Dekan

Share this post


Link to post
Share on other sites

normally if there is not specification about + or - you should mention it as + or -4.

 

Or depending on the context, e.g. in basic geometry, one would typically use the principal value, since the negative is intuitively meaningless for describing length, area, etc.

Share this post


Link to post
Share on other sites

Both the forms are correct.

 

Consider

 

squareroot(16) = x

16 = x square

so if both - and + comes for x it will be always a positive number

 

i.e 16 = > (-4) square =16 and 4 (square) = 16

 

So √16=±4 is correct.

Share this post


Link to post
Share on other sites

Both the forms are correct.

 

Consider

 

squareroot(16) = x

16 = x square

so if both - and + comes for x it will be always a positive number

 

i.e 16 = > (-4) square =16 and 4 (square) = 16

 

So √16=±4 is correct.

 

Not really.

 

16 does has two distinct square roots, as does every other real number (besides zero, which has only itself). Nothing will change that. However, notation is not flexible and we denote specific things in specific ways according to convention. The radical [math]\sqrt{n}[/math] almost always denotes only the principal value.

 

The following statements are true according to the typical convention.

 

[math]\sqrt{16}=4[/math]

[math]\sqrt{16}\ne \pm 4[/math]

[math]\pm\sqrt{16}=\pm 4[/math]

 

The latter of the three is how one should express both the positive and negative values.

Edited by Amaton

Share this post


Link to post
Share on other sites

 

Not really.

 

16 does has two distinct square roots, as does every other real number (besides zero, which has only itself). Nothing will change that. However, notation is not flexible and we denote specific things in specific ways according to convention. The radical [math]\sqrt{n}[/math] almost always denotes only the principal value.

 

The following statements are true according to the typical convention.

 

[math]\sqrt{16}=4[/math]

[math]\sqrt{16}\ne \pm 4[/math]

[math]\pm\sqrt{16}=\pm 4[/math]

 

The latter of the three is how one should express both the positive and negative values.

The reason why there are two solutions, the positive and negative, is because if you square a negative number it becomes positive. Therefore, it would be correct to say that [math]\sqrt{16}= \pm 4[/math].

 

Typical convention says that it equals both until specified.

Edited by Unity+

Share this post


Link to post
Share on other sites

The notation [math]\sqrt{x}[/math], for [math]x\in\mathbb{R}_{\ge 0}[/math], refers to the principal square root (which is positive) unless otherwise specified. Amaton's post is correct (except that notation is somewhat flexible--it just depends on how much you want to annoy your readers tongue.png).

Edited by John

Share this post


Link to post
Share on other sites

The notation [math]\sqrt{x}[/math], for [math]x\in\mathbb{R}_{\ge 0}[/math], refers to the principal square root (which is positive) unless otherwise specified. Amaton's post is correct (except that notation is somewhat flexible--it just depends on how much you want to annoy your readers tongue.png).

True, but I didn't see anything referring to what values were acceptable. happy.png

Share this post


Link to post
Share on other sites

 

The notation 1a03d0f7242823c05e0f16ad19f85201-1.png, for 00f691c1c562200ce5f363adc040bb17-1.png, refers to the principal square root (which is positive) unless otherwise specified. Amaton's post is correct (except that notation is somewhat flexible--it just depends on how much you want to annoy your readers tongue.png).

 

I would dispute that your notation is the same as previously posted.

 

The square root function does indeed follow your definition by convention.

 

However the equation (which is not a function) must allow any value that satisfies the equality, by definition.

 

It is important to distinguish between functions and expressions containing an equals sign.

 

When we have an expression containing an equals sign that may be satisfied by more than one value we would normally have additional conditions to satisfy which will determine which one we want.

 

There were several mistakes mades in a recent thread on quadratics by posters failing to observe this simple requirement.

 

So John's statement is a refers to a set of values for x (the domain) and another set of values that picks out a unique member for each value of x.

 

That properly makes it a function.

 

Amaton referred to a specific value of x (16) and presented it via an equals sign.

 

That does not qualify as a function.

Edited by studiot

Share this post


Link to post
Share on other sites

There are two square roots of 16, but [math]\sqrt{16} = 4[/math] unless otherwise specified. In general, [math]\forall x \in \mathbb{R}, \sqrt{x^{2}} = |x|[/math]. If we have the equation [math]x^{2} - 16 = 0[/math], then [math]x = \pm \sqrt{16} = \pm 4[/math]. It's not a huge deal, regardless, but Amaton's post is correct.

Edited by John

Share this post


Link to post
Share on other sites

 

 

There are two square roots of 16, but [math]\sqrt{16} = 4[/math] unless otherwise specified. In general, [math]\forall x \in \mathbb{R}, \sqrt{x^{2}} = |x|[/math]. If we have the equation [math]x^{2} - 16 = 0[/math], then [math]x = \pm \sqrt{16} = \pm 4[/math]. It's not a huge deal, regardless, but Amaton's post is correct.

 

 

I didn't agree with you before and I don't agree with you now.

 

Neither have you addressed my comments.

 

Take for instance the equation x2+3x = -2

 

Would you bar the negative roots?

Share this post


Link to post
Share on other sites

Er, no, why? Finding the roots of a general polynomial equation isn't the same as taking the square root of a real number. And as stated before, if you look at [math]x^{2} - 16 = 0[/math], then there are two solutions, [math]\sqrt{16} = 4[/math] and [math]-\sqrt{16} = -4[/math]. But the notation, [math]\sqrt{16}[/math], refers to the principal square root of 16, i.e. 4. Disagree if you like, but that is the convention.

Share this post


Link to post
Share on other sites

I don't agree with your convention and you still haven't addressed my point that this is because you are talking about a function not an equation.

 

What would you make of this square root ?

 

[math]\sqrt {\left( {5 + \sqrt {21} } \right)} [/math]

 

Which again is not a function

 

or perhaps this one, which is

In particular can you guarantee that it is always positive ?

 

[math]\sqrt {\left( {1 - 2x + 5{x^2} - 4{x^3} + 4{x^4}} \right)} [/math]


Here are some more thoughts.

 

Do you consider

 

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

To be an identity?

 

If so do you agree that

 

[math]\sqrt {{{\left( { - 4} \right)}^2}} [/math][math] = [/math][math] - 4[/math]


Edited by studiot

Share this post


Link to post
Share on other sites

It's not my convention. It's the convention used by mathematicians in general. While the fact that it's a convention means there's probably no absolute authority to make the definition, a Google search for definitions of the square root or the radical sign should convince you.

And I'm honestly not sure what your point is. All I've said is that [math]\forall x \in \mathbb{R}, \sqrt{x^{2}} = |x|[/math]. It's simply a matter of notation. Questions about the roots of various polynomial equations are somewhat unrelated.

But for the particular example of [math]\sqrt {\left( {1 - 2x + 5{x^2} - 4{x^3} + 4{x^4}} \right)}[/math], so long as [math]1 - 2x + 5{x^2} - 4{x^3} + 4{x^4} > 0[/math], then yes, [math]\sqrt {\left( {1 - 2x + 5{x^2} - 4{x^3} + 4{x^4}} \right)} > 0[/math].

Using Wolfram Alpha, it seems the two square roots of [math]1 - 2x + 5{x^2} - 4{x^3} + 4{x^4}[/math] (which, as it turns out, is positive for all real x) are [math]\pm(2x^2 - x + 1)[/math], so [math]\sqrt {\left( {1 - 2x + 5{x^2} - 4{x^3} + 4{x^4}} \right)} = |2x^2 - x + 1|[/math], which is positive.

Edit: Just saw your edit. The notation indicates the principal square root, so [math]\sqrt{(-4)^{2}} = |{-4}| = 4[/math].

Edited by John

Share this post


Link to post
Share on other sites

 

John post#18

And I'm honestly not sure what your point is.

 

Perhaps that's because you have still not acknowledged what I said several times about the difference between functions and equations (or expressions containg equations).

 

Studiot post#13

It is important to distinguish between functions and expressions containing an equals sign.

 

Incidentally I'm sorry the material I added whilst you were thinking was tagged onto my previous post by the system. I genuinely tried to make a new post.

Share this post


Link to post
Share on other sites

The difference between an equation and a function is irrelevant. The notation [math]\sqrt{x}[/math] refers to the principal square root of x. It's not strange. It's not controversial. I don't know why this is such a sticking point.

Edited by John

Share this post


Link to post
Share on other sites

 

The difference between an equation and a function is irrelevant.

 

Why do you say this?

 

IMHO it is of the essence of the issue.

Share this post


Link to post
Share on other sites

We're just talking about the definition of the radical sign. By default, it indicates the principal (i.e. positive) square root of the radicand. In situations where both square roots of the radicand are desired, we prepend [math]\pm[/math] to specify this. So if someone asks for [math]\sqrt{16}[/math], i.e. "the square root of 16," the answer is 4. If someone asks for solutions to the equation [math]x^2 = 16[/math], i.e. "what numbers are square roots of 16," the answers are 4 and -4, and we can express these as [math]\pm\sqrt{16}[/math]. If you want to use [math]\sqrt{16}[/math] to mean both 4 and -4, then that's fine, but it may cause some confusion unless this usage is specified (though often the intended meaning will probably be clear from the context).

Share this post


Link to post
Share on other sites

You keep making statements as absolute, without any backing or logical development.

 

For instance you keep appealing to mathematical convention.

 

Here is an excerpt from the most successful and long lived textbook on the subject ever written in English (and translated into umpteen languages) which has survived for more than a century as the gold standard

 

 

116 Definition The root of any proposed expression is that quantity which being multiplied by itself the requisite number of times produces the given expression.

 

The operation of finding the root is called Evolution. It is the reverse of Involution.

 

117 By the Rule of signs we see that

(1) any even root of a positive quantity may be either positive or negative;

!2) no negative quantity can have an even root

(3)every odd root of a quantity has the same sign as the quantity itself

 

Has someone recently proved messers Hall & Knight wrong in that last century?

I have not heard of it.

Share this post


Link to post
Share on other sites

I'm stating the (usual) definition of notation. You might as well ask me to logically develop why we normally use "+" to denote addition. If you did, then I'd say the same thing: It's a matter of convention, and a Google search should provide sufficient evidence to show this.

 

Notation isn't sacred. If I like, I can define [math]\sqrt{x}[/math] to refer to [math]e^{x}[/math] or [math]x \choose {x-6}[/math] or anything else. It'd be odd, and there's no good reason--or at least, no reason that outweighs the confusion it'd likely cause--to do so, but I could. However, the usual definition of [math]\sqrt{x}[/math] is the principal square root of x. This doesn't require rigorous development. It's simply the way notation has developed over the years.

This definition also doesn't conflict with the quoted passage, unless you're referring to the fact that I translated [math]\sqrt{x}[/math] as "the square root of x." But if that bothers you, then ignore it. The rest of my post remains true.

Edited by John

Share this post


Link to post
Share on other sites

Well actually it said any even root may be positive or negative. Does that not include the square root?

Edited by studiot

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
Sign in to follow this  

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.