- #1

ShayanJ

Gold Member

- 2,809

- 604

Hi

Can someone give me a proof for sqrt(x^2)=|x|?

I mean why sqrt(x^2)!=+x,-x?

thanks

Can someone give me a proof for sqrt(x^2)=|x|?

I mean why sqrt(x^2)!=+x,-x?

thanks

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter ShayanJ
- Start date

- #1

ShayanJ

Gold Member

- 2,809

- 604

Hi

Can someone give me a proof for sqrt(x^2)=|x|?

I mean why sqrt(x^2)!=+x,-x?

thanks

Can someone give me a proof for sqrt(x^2)=|x|?

I mean why sqrt(x^2)!=+x,-x?

thanks

- #2

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 969

It is true that, for example, [itex]x^2= 4[/itex] has two roots: x= -2 and x= 2. But [itex]\sqrt{4}[/itex] is specifically defined as "the

In fact, the reason

- #3

- 2,328

- 309

[tex]f(x)=x^2[/tex]

is not

[tex] y = x^2[/tex]

To be invertible the inverse relation must be a

In order to "make" our square function invertible we must restrict the domain:

[tex]f_+(x) = x^2\quad (x\ge 0)[/tex]

(i.e. if x is negative the function is undefined)

In this case we can then define an inverse function

[tex] f_+^{-1}(y) = \sqrt{y}[/tex]

Note we could have chosen to restrict the domain differently:

[tex] f_-(x) = x^2 \quad (x \le 0)[/tex]

This then would have inverse:

[tex] f_-^{-1}(y)= - \sqrt{y}[/tex]

The square root function gives the "principle root". Since there are two solutions to the equation x^2 = 4, namely 2 and -2 then in order that sqrt be a function (have a unique value) we must pick one of the two solutions. We adopt the convention of choosing the positive solution and call this the "principle" solution. We could have "gone the other way" but more often the solution we want is the positive one.

As for your proof you should break the problem down into cases, x = 0, x>0, x < 0. Show that the value output is the same as |x| in each case.

- #4

- 290

- 2

How about

[tex]|z|=|a+bi|=\sqrt{a^2+b^2}\implies |a|=\sqrt{a^2}[/tex]

[tex]|z|=|a+bi|=\sqrt{a^2+b^2}\implies |a|=\sqrt{a^2}[/tex]

- #5

- 27

- 1

It's more of a definition than anything you can prove.

- #6

ShayanJ

Gold Member

- 2,809

- 604

And to you tibarn.What if I tell you that the sentence "There are infinite number of prime numbers.",has a proof?

- #7

matt grime

Science Advisor

Homework Helper

- 9,420

- 4

I've no idea what you mean by your last sentence, Shyan. It is the definition of the square root function (or perhaps convention if you prefer) that the square root of a positive number r is the positive one of the two possible solutions to x^2=r. We could form a perfectly good version of mathematics where it is the negative one. It would be perverse and silly but perfectly possible. In that sense you can't 'prove' it in any non-trivial way

- #8

ShayanJ

Gold Member

- 2,809

- 604

Thank every one for the answers

- #9

- 302

- 0

Because it is DEFINED that way. Same thing goes for the square root. "qntty" has not proved anything.

- #10

ShayanJ

Gold Member

- 2,809

- 604

first:|x|=x if x>=0,-x if x<0

second:Yes he did.

sqrt(x^2) is defined to be |x|.but I didn't know why it is.qntty told the reason.So he PROVED it.

In fact he passed the way that the mathematicians has passed to reach to the definition of sqrt(x^2) in front of our eyes.And I think that is one kind of PROVING.

second:Yes he did.

sqrt(x^2) is defined to be |x|.but I didn't know why it is.qntty told the reason.So he PROVED it.

In fact he passed the way that the mathematicians has passed to reach to the definition of sqrt(x^2) in front of our eyes.And I think that is one kind of PROVING.

Last edited:

- #11

- 302

- 0

NO NO NO. You are using the definition and that is no good according to your own standards. PROVE it to me!

- #12

matt grime

Science Advisor

Homework Helper

- 9,420

- 4

- #13

ShayanJ

Gold Member

- 2,809

- 604

if it was a definition too:

then there are just two candidates for the definition of |x|.yours and mine.So if I prove that yours is wrong,then mine will be true.And that's exactly what I'm going to do.

Imagine x=-2.as your definition says,|-2|=-2,but we all know that a modulus won't be equal to a negative number.So your definition is wrong and mine is true.

- #14

- 302

- 0

Now prove that modulus is always positive.

- #15

ShayanJ

Gold Member

- 2,809

- 604

Is it finished?

- #16

- 494

- 0

If I ask "What is sqrt(25)?", the answer is +5.

If I ask "Which real number x has the property that x^2 = 25?", the answer is there are two: +5 and -5.

sqrt(25) is defined to be a function, hence, single-valued.The positive root is taken, as far as I can tell, because writing it takes less time (the + sign can be assumed on positive numbers...)

- #17

matt grime

Science Advisor

Homework Helper

- 9,420

- 4

You are not seeking a proof that is anything other than the vacuous one. You might perhaps want a justification for it. The functions |x| and sqrt(x^2) are identical on the real numbers, and that is an immediate consequence of the definition of each. It is not a deep fact, and it requires nothing that really merits the word 'proof'.

- #18

ShayanJ

Gold Member

- 2,809

- 604

And that's because we can fight for ever and get no result.so I think you are right at the end.and sqrt(x^2)=|x| is sth like an axiom or maybe a law.

Ok

thanks to all.

- #19

- 27

- 1

|x| equals x if x>0 and -x otherwise.

Therefore, sqrt(x^2) = |x|.

Pedantic, yes, but this is a definition, not a deep fact.

- #20

- 302

- 0

An example: By definition, the word "knife" refers to a metallic or plastic object used for cutting stuff. You can use it to stab someone or you can use it when you eat. That is how "knife" is defined. IT IS COMPLETELY ARBITRARY. You might as well define "knife" to mean the same as "monkey" or whatever. Go ahead, invent your own language.

The same thing goes for the symbol "sqrt". Is has a certain definition, a certain meaning.

Share: