# Definition of absolute value

1. Oct 25, 2012

### stauros

Which of the following two definitions is correct:

1) $\forall x\forall y[ |x|=y\Longleftrightarrow( x\geq 0\Longrightarrow x=y)\wedge(x<0\Longrightarrow x=-y)]$

2) $\forall x\forall y[ |x|=y\Longleftrightarrow( x\geq 0\Longrightarrow x=y)\vee(x<0\Longrightarrow x=-y)]$

I think the second one

2. Oct 25, 2012

### Erland

No, the first one is correct, and the second one is wrong.

Recall that $A\Longrightarrow B$ is true if A is false, and also if B is true. It is false if and only if A is true and B is false.

Therefore $(x\geq 0\Longrightarrow x=y)\vee(x<0\Longrightarrow x=-y)$ is true no matter which values x and y have (as long as they are real numbers). For if $x\geq 0$, then the second implication is automatically true, and if $x<0$, then the first implication is true. Thus, the disjunction is always true, so 2) then means that |x|=y for all y, which makes no sense.

But for 1), if $x\geq 0$, then the second implication is automatically true, while the first one is true only for x=y, and thus the conjunction is true only if x=y. If, instead, $x<0$, then the first implication is automatically true, while the second one is true only for x=-y, and thus the conjunction is true only if x=-y. Together, this means that the conjunction is true if and only if |x|=y, just as we desired.

3. Oct 25, 2012

### stauros

But also the second definition,to use your words," is true no matter which values x and y have"

Because :

$[(x\geq 0\Longrightarrow x=y)\wedge(x<0\Longrightarrow x=-y)]\Longrightarrow [(x\geq 0\wedge x<0)\Longrightarrow(x=y\wedge x=-y)]$

Which is always true ,since $(x\geq 0\wedge x<0)$ is always false

Last edited: Oct 25, 2012
4. Oct 26, 2012

### Stephen Tashi

The second definition you gave uses $\vee$, not $\wedge$.

Showing that statement $(x \geq 0 \Rightarrow x = y) \wedge (x \lt 0 \Rightarrow x = -y)$ implies a true statement for each value of x and y does not show that $(x \geq 0 \Rightarrow x = y) \wedge (x \lt 0 \Rightarrow x = -y)$ itself is a true statement for all values of x and y.

The statement $( x \ge 0 \Rightarrow x = y) \vee (x \lt 0 \Rightarrow x = -y)$
is true for all values of x and y. Hence it is true for x = 6 and y = -6. If the statement were equivalent to the statement $|x| = y$ then |6| = -6 would be true.

The statement $(x \ge 0 \Rightarrow x = y) \wedge (x \lt 0 \Rightarrow x = -y)$ is not true for all values of x and y. It is not true for x = 6 and y =-6 because the statement $(x \ge 0 \Rightarrow x = y)$ is not true.

5. Oct 26, 2012

### stauros

Sorry my mistake. I should have reffered to the 1st definition instead of the 2nd one.

Your thinking looks quite reasonable ,but then again ,if the 1st definition is correct we should be able to get:

$\forall x[ (x\geq 0\Longrightarrow |x|=x)\wedge(x<0\Longrightarrow |x|=-x)]$ instead of:

$\forall x[ (x\geq 0\Longrightarrow |x|=x)\vee(x<0\Longrightarrow |x|=-x)]$

6. Oct 26, 2012

### Erland

Indeed. This is true and it follows from 1).

The statement

$\forall x[ (x\geq 0\Longrightarrow |x|=x)\vee(x<0\Longrightarrow |x|=-x)]$

is also true, but not interesting wrt the definition of absolute value.

7. Oct 26, 2012

### stauros

How ??

8. Oct 27, 2012

### Erland

Assume that 1) is true. This means that |x|=y⟺(x≥0⟹x=y)∧(x<0⟹x=−y)

holds for all x and y, including all x and y such that y=|x|, so

|x|=|x|⟺(x≥0⟹x=|x|)∧(x<0⟹x=−|x|)

for all x. But |x|=|x| is certainly true, so

(x≥0⟹x=|x|)∧(x<0⟹x=−|x|)

holds for all x, which is the same as

∀x[(x≥0⟹|x|=x)∧(x<0⟹|x|=−x)].

9. Oct 27, 2012

### stauros

This does not look very convincing to me, how can you know that what you have proved is correct

10. Oct 27, 2012

### Erland

It is a chain of logical reasoning. Which steps do you doubt?

11. Oct 28, 2012

### stauros

exactly ,i am confused with the logical reasoning. What is logical reasoning anyway

12. Oct 28, 2012

### Stephen Tashi

Where did you get the question in your original post? Aren't you studying some materials that deal with logic?

13. Oct 28, 2012

### stauros

In the "axiomatic set theory" book of Patrick Suppes ,page 172

14. Oct 28, 2012

### Erland

Stauros, just tell us which steps in the above reasoning (in Post 8) you have difficulties with. Otherwise, it is hard to help you.

15. Oct 28, 2012

### stauros

O.k ,in substituting in the definition y =|x| are you allowed to do that?

And then how do you jump from :

$|x|=|x|\Longleftrightarrow (x\geq 0\Longrightarrow x=y)\wedge (x<0\Longrightarrow x= -y)$ to

$(x\geq 0\Longrightarrow |x|=x)\wedge (x<0\Longrightarrow |x|= -x)$

16. Oct 29, 2012

### Erland

Yes. $\forall x\forall y P(x,y)$ means that $P(x,y)$ holds no matter which values x and y have. It must then also hold for all values of x and y such that y=|x|. Therefore $P(x,|x|)$ must hold for all x.

As an analogous example, you agree that the rule

$\forall x\forall y[(x+y)(x-y)=x^2-y^2]$

holds, right?

Then, don't you also agree that e.g.

$(x+\sin x)(x-\sin x)=x^2-\sin^2 x$

holds for all x, and that this is a consequence of the rule?
That $A \Longleftrightarrow B$ means that $A$ and $B$ either are both true or both false. Then, if we now that $A \Longleftrightarrow B$ is true and that also $A$ is true, then $B$ must also be true.

Since we established that

$|x|=|x|\Longleftrightarrow (x\geq 0\Longrightarrow x=|x|)\wedge (x<0\Longrightarrow x= -|x|)$

(you wrote it erroneously, with y instead of |x|) is true for all x, and since, certainly, $|x|=|x|$ is true for all x, we obtain that

$(x\geq 0\Longrightarrow x=|x|)\wedge (x<0\Longrightarrow x= -|x|)$

holds for all x.

17. Oct 30, 2012

### stauros

How about if you put y=x,then you have :

$[|x|=x\Longleftrightarrow (x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)]$

Is that true ,and if yes how do we prove it?

1) I think you have not established that :

$[|x|=|x|\Longleftrightarrow (x\geq 0\Longrightarrow x=|x|)\wedge(x<0\Longrightarrow x=-|x|)]$ is true,for all values of x

You have just assume that the above holds for a particular value of x and y.

To establish that the above holds for all values of x and y you must try all real values of x and y

2) Why $|x|= |x|$ is always true for all x?

18. Oct 30, 2012

### Erland

Yes, it is true. It also follows from

1) $\forall x\forall y[ |x|=y\Longleftrightarrow( x\geq 0\Longrightarrow x=y)\wedge(x<0\Longrightarrow x=-y)]$.

Of course, if you don't accept 1), you don't have to accept its consequences either (at least not without more facts). But if we accept 1), which we should do since it can be taken as a definition of absolute value (it precisely expresses what we mean with absolute value), then we must also accept its logical consequences, such as (*) and

$[|x|=|x|\Longleftrightarrow (x\geq 0\Longrightarrow x=|x|)\wedge(x<0\Longrightarrow x=-|x|)]$

(both true for all (x)).

However, it is also easy to verify (*) directly, without deducing it from 1). I leave this as as an exercise for you. (Hint: Verify that if $x\ge 0$ then both sides of the equivalence in (*) are true, and if $x<0$ then both sides are false. This means that the equivalence is true for all values of $x$.)
Yes, I established that it follows from 1). Nothing more is needed. But of course, one can verify also this without using 1). You can try that.
How can you doubt this? Do you really believe that something can be unequal to itself?

19. Oct 31, 2012

### stauros

1) First of vall i think you agree that verification is not a proof

2)Your hint inspired me for the following proof :

Let $|x|=x$............................................................................1

We know from the law of trichotomy that:

$x\geq 0\vee \ x<0$................................................................2

Let : $x\geq 0$..........................................................................3

But we know that : $x=x$.............................................................4

Hence : $x\geq 0\Longrightarrow x=x$............................................5

Let : $x<0$..................................................................................6

But we know that : $x<0\Longrightarrow |x|=-x$...............................7

Hence from (6) and(7) we conclude that : $|x|=-x$..............................................8

And by substituting (1) into (8) we have: $x=-x$..................................................9

Thus : $x<0\Longrightarrow x=-x$......................................................................10

And by the power of the law of logic called proof by cases we can conclude:

$(x\geq 0\Longrightarrow x=x)\vee(x<0\Longrightarrow x=-x)$

Therefor : $|x|=x\Longrightarrow (x\geq 0\Longrightarrow x=x)\vee(x<0\Longrightarrow x=-x)$

So:

$|x|=x\Longrightarrow (x\geq 0\Longrightarrow x=x)\vee(x<0\Longrightarrow x=-x)$

is right ,and

$|x|=x\Longrightarrow (x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)$

is wrong

Hence:

$|x|=x\Longleftrightarrow (x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)$

is not true

20. Oct 31, 2012

### Erland

That is in itself correct. For a correct proof, 1) should be used. But I had in mind a more direct argument, where we freely use these two properties of absolute value:

$x\ge 0 \Longrightarrow |x|=x$ and $x<0 \Longrightarrow |x|=-x$ (the last one you also used in your dervation, step 7).

These properties are actually consequences of 1), and in a rigorous proof of (*) it is unnecessary to prove and use these, for we only need to set y=x. But still it can be instructive to use these two properties in a proof (or argument) to better understand why (*) is true.

Everything in the derivation is correct thus far, but then you claim:
This does not follow. What you claim is wrong is not wrong, but right.

Instead, you can do like this:

Assume first that $x\ge 0$. Prove that then $|x|=x$ is true and

$(x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)$

is true. Conclude that then is

$|x|=x\Longleftrightarrow(x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)$

true.

Assume then that $x<0$. Prove that then $|x|=x$ is false and

$(x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)$

is false. Conclude that then is

$|x|=x\Longleftrightarrow(x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)$

true.

Conclude that in all cases (for all x)

$|x|=x\Longleftrightarrow(x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)$

is true.