# How would you refute this argumentation?

1. Nov 9, 2004

### quasar987

It is true for all x real that

$$|x|^2 = x^2 \Leftrightarrow \sqrt{|x|^2} = \sqrt{x^2} \Leftrightarrow |x|^{2/2} = x^{2/2} \Leftrightarrow |x| = x$$

for all x real, which is not true.

This is not an assignment I have, I'm just wondering. Thanks.

2. Nov 9, 2004

### Zurtex

$$\sqrt{|x|^2} = \sqrt{x^2} \Leftrightarrow |x|^{2/2} = x^{2/2}$$

This is where is breaks down, it is not true the right hand side of these 2 equalities are the same. If you are taking the square root symbol to be one to one then you are only defining it for a positive co-domain, however:

$$x^{\frac{2}{2}}$$

Is defined for more than a positive co-domain. I've probably explained myself really badly being 10 to 3 in the mourning where I live but I hope that helps.

3. Nov 10, 2004

### Galileo

The square root of $x^2$ is not equal to x, but |x| !

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

4. Nov 15, 2004

### James R

For real x:

$$|x|^2 = x^2$$

True

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

True.

$$|x|^{2/2} = x^{2/2}$$

False, since it implies

$$|x| = x$$

which is only true for x &gt; 0.

5. Nov 15, 2004

### gerben

For real x:

$$|x|^2 = x^2$$

True

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

True.

$$|x|^{2/2} = x^{2/2}$$

False, since
$$\sqrt{|x|^2} \neq |x|$$

$$\sqrt{|x|^2} = x \vee -x$$

6. Nov 15, 2004

### jcsd

gerben $\sqrt{x} = |x| \therefore \sqrt{|x|^2} = |x|$

7. Nov 16, 2004

### gerben

jcsd, I do not see what you mean (especially not what you mean with the squareroot of x being equal to the absolute value of x).

I was using:
$$\sqrt{a^2} = \pm a$$
so if
$$a = |x|$$
$$\pm |x|$$
which is the same as
$$\pm x$$

8. Nov 16, 2004

### jcsd

Soory that should of been $\sqrt{x} = |\sqrt{x}|$ (which necessarily means that $\sqrt{x}$ is always postive).

9. Nov 16, 2004

### philosophking

I think I can clear this up. Your first case is true, but you have to do it conditionally. This is what I mean:

|x| = x if x>0
|x| = -x if x<0

THEREFORE

|x|^2 = (x)^2 if x>0
|x|^2 = (-x)^2 if x<0

et cetera

10. Nov 17, 2004

### gerben

Well, that depends on your definition of square root, I use a definition like this:
"the number that when multiplied by itself will produce a given number"

When looking at matworld (http://mathworld.wolfram.com/SquareRoot.html) I also found:
"Note that any positive real number has two square roots, one positive and one negative."

...

I think that with your definition you cannot refute the derivation in the original post, which leads to a result that you do not want to accept, so somewhere you have to change a rule or definition. Perhaps you see another possibility?

11. Nov 17, 2004

### jcsd

No my defintion is the correct one $\sqrt{x}$ is always postive as that is simply it's definition, that should be clear from your mathworld link.

As for the problem the solution is simple:

$$|x|^{\frac{2}{2}} \ne x^{\frac{2}{2}}$$

12. Nov 17, 2004

### gerben

No if you follow the mathwolrd link you see that the square root is defined as:
"A square root, also called a radical or surd, of x is a number r such that $r^2 = x$"
so there are two solutions a positive one and a negative one, the positive solution is also called "the principal square root"... yeah well just take a look for yourself...

As you can see in post #5 I agree that this is the point where a rule is broken, and I have explained which rule. If you do not agree with this, you have to come up with something else that is wrong, otherwise you have rules that can prove something that you do not agree with.

The problem was that by following seemingly valid rules you would arrive at an invalid expression, so somewhere you must have followed an invalid rule, where is it and which rule? Just saying that the expression is false does not explain anything.

13. Nov 18, 2004

### Zlex

$$\sqrt{x}$$ is always positive?

Counter:

$$x^2 = a$$, condition: a > 0
$$x = \sqrt{a}$$
$$(\sqrt{a})^2 = a$$

But,
$$(-\sqrt{a})^2 = a$$

Eg. $$(-2)^2 = 4[tex] Thus root of 4 is +- 2 14. Nov 18, 2004 ### jcsd But where did I say I was talking about the term 'square root'? I said $\sqrt{x}$ is always postive that's because this sign signifies the principal square root, $-\sqrt{x}$ signifies the neagtive square root of a real number. This is clearly stated in the Mathworld link. The whole reason why the 'argumenatation' is wrong i s because it doen't distinguish between the principal value and th eother value sof the square root. 15. Nov 18, 2004 ### jcsd It's a definition; $\sqrt{x}$ is simply defined as the postive square root. Last edited: Nov 18, 2004 16. Nov 18, 2004 ### gerben Ok I see, you are right. Yes, and apparently the "other way around" than I thought: $\sqrt{x} = |x|$ (because $\sqrt{-x^2} = \sqrt{x^2} = x$) so the argument in the original post should be: [tex]\sqrt{|x|^2} = \sqrt{x^2} \Leftrightarrow |x|^{2/2} = |x^{2/2}| \Leftrightarrow |x| = |x|$$

17. Nov 19, 2004

### Alkatran

I'm surprised how common equations like this one are.

Basic steps to this:
Find a function which can be many to one (such as ^2)
Start from the first 'many' element (-2, for example)
Use the function on the value (get 4)
Use the inverse but use the second 'many' element (get 2)
OMG 2 == -2 !!!!

This is the kind of logic that makes you think functions need to return context along with numbers. (comes from positive branch)