## Definition of absolute value

 Quote by 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?
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##.)
 Quote by stauros 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.
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.
 Quote by stauros 2) Why ## |x|= |x|## is always true for all x?
How can you doubt this? Do you really believe that something can be unequal to itself?

 Quote by Erland 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##.)
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

 Quote by stauros 1) First of vall i think you agree that verification is not a proof
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.

 Quote by stauros 2)Your hint inspired me for the following proof : [---] So: ##|x|=x\Longrightarrow (x\geq 0\Longrightarrow x=x)\vee(x<0\Longrightarrow x=-x)## is right
Everything in the derivation is correct thus far, but then you claim:
 Quote by stauros ##|x|=x\Longrightarrow(x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)## is wrong.
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.

 Quote by Erland 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. .
How??

 Assume that ##x\ge 0##. Then ##|x|=x##. Since ##x=x##, we also have ##x\geq 0\Longrightarrow x=x##. Since ##x\ge 0##, ##\neg \, x<0##, which gives ##x<0\Longrightarrow x=-x##. Thus, ##(x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)##. But we also had ##|x|=x##, so ##|x|=x\Longleftrightarrow(x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)## (if we assume ##x\ge 0##, that is).

 Quote by Erland Assume that ##x\ge 0##. Then ##|x|=x##. Since ##x=x##, we also have ##x\geq 0\Longrightarrow x=x##. Since ##x\ge 0##, ##\neg \, x<0##, which gives ##x<0\Longrightarrow x=-x##. Thus, ##(x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)##. But we also had ##|x|=x##, so ##|x|=x\Longleftrightarrow(x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)## (if we assume ##x\ge 0##, that is).
Sorry ,i cannot follow.

Can you write your proof in a stepwise manner in more details ,like i did mine?

So i will not have to ask questions for each line of the proof.

Thanks

 Quote by stauros Sorry ,i cannot follow. Can you write your proof in a stepwise manner in more details ,like i did mine? So i will not have to ask questions for each line of the proof. Thanks
I think it is better to make a partial truth table, for all combinations of truth values of the atomic formulae that can occur together. A full truth table has ##2^5=32## rows, but it turns out that we need only check 3 of them, for the other 29 combinations of truth values can never occur together in this case.

The only three combinations that can occur together are:

1)
a) ##|x|=x## true
b) ##x\ge 0## true
c) ##x=x## true
d) ##x<0## false
e) ##x=-x## true
(this holds if and only if ##x=0##, the only case where ##x=-x## is true)

2)
a) ##|x|=x## true
b) ##x\ge 0## true
c) ##x=x## true
d) ##x<0## false
e) ##x=-x## false
(this holds if and only if ##x>0##)

3)
a) ##|x|=x## false
b) ##x\ge 0## false
c) ##x=x## true
d) ##x<0## true
e) ##x=-x## false
(this holds if and only if ##x<0##)

You should convince yourself that these are the only combinations of truth values of these five atomic formulae that can occur.

Now, if we use the truth tables for implication, conjunction, and equivalence, we see that:

Case 1) and 2):
i) ##|x|=x## true (1 a or 2 a)
ii) ##x\ge 0\Longrightarrow x=x## true (from 1 b and c, or 2 b and c)
iii) ##x<0\Longrightarrow x=-x## true (from 1 d and e, or 2 d and e)
iv) ##(x\ge 0\Longrightarrow x=x)\wedge (x<0\Longrightarrow x=-x)## true (from ii and iii)
v) ##|x|=x\Longleftrightarrow(x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)## true (from i and iv)

Case 3):
i) ##|x|=x## false (3 a)
ii) ##x\ge 0\Longrightarrow x=x## true (from 3 b and c)
iii) ##x<0\Longrightarrow x=-x## false (from 3 d and e)
iv) ##(x\ge 0\Longrightarrow x=x)\wedge (x<0\Longrightarrow x=-x)## false (from ii and iii)
v) ##|x|=x\Longleftrightarrow(x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)## true (from i and iv)

Thus, in all cases which can occur: ##| x|=x\Longleftrightarrow(x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)## is true, so this formula is true for all ##x##.

In other words:
##\forall x[|x|=x\Longleftrightarrow(x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)]##
is true.

 Quote by Erland I think it is better to make a partial truth table, for all combinations of truth values of the atomic formulae that can occur together. A full truth table has ##2^5=32## rows, but it turns out that we need only check 3 of them, for the other 29 combinations of truth values can never occur together in this case. The only three combinations that can occur together are: 1) a) ##|x|=x## true b) ##x\ge 0## true c) ##x=x## true d) ##x<0## false e) ##x=-x## true (this holds if and only if ##x=0##, the only case where ##x=-x## is true) 2) a) ##|x|=x## true b) ##x\ge 0## true c) ##x=x## true d) ##x<0## false e) ##x=-x## false (this holds if and only if ##x>0##) 3) a) ##|x|=x## false b) ##x\ge 0## false c) ##x=x## true d) ##x<0## true e) ##x=-x## false (this holds if and only if ##x<0##) You should convince yourself that these are the only combinations of truth values of these five atomic formulae that can occur. Now, if we use the truth tables for implication, conjunction, and equivalence, we see that: Case 1) and 2): i) ##|x|=x## true (1 a or 2 a) ii) ##x\ge 0\Longrightarrow x=x## true (from 1 b and c, or 2 b and c) iii) ##x<0\Longrightarrow x=-x## true (from 1 d and e, or 2 d and e) iv) ##(x\ge 0\Longrightarrow x=x)\wedge (x<0\Longrightarrow x=-x)## true (from ii and iii) v) ##|x|=x\Longleftrightarrow(x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)## true (from i and iv) Case 3): i) ##|x|=x## false (3 a) ii) ##x\ge 0\Longrightarrow x=x## true (from 3 b and c) iii) ##x<0\Longrightarrow x=-x## false (from 3 d and e) iv) ##(x\ge 0\Longrightarrow x=x)\wedge (x<0\Longrightarrow x=-x)## false (from ii and iii) v) ##|x|=x\Longleftrightarrow(x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)## true (from i and iv) Thus, in all cases which can occur: ##| x|=x\Longleftrightarrow(x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)## is true, so this formula is true for all ##x##. In other words: ##\forall x[|x|=x\Longleftrightarrow(x\geq 0\Longrightarrow x=x)\wedge(x<0\Longrightarrow x=-x)]## is true.
Before we examine the correctness of your proof,

Can you use your semantical method of true and false values to find out whether the formula:

##\forall a[a\neq 0\Longrightarrow a^2>0]## is a theorem or not?

 Quote by stauros Before we examine the correctness of your proof, Can you use your semantical method of true and false values to find out whether the formula: ##\forall a[a\neq 0\Longrightarrow a^2>0]## is a theorem or not?
If you want to prove that something is a theorem, you should make a derivation from the axioms. My intention with the "semantic" method was not to prove that the sentence in question is a theorem, but to convince you that it is true, so that the proposed definition of absolute value does not lead to something false. The sentence in the previous post can be derived in a few steps from the given definition of absolute value, by using logical axioms and rules of inference (exactly how depends on which logical axioms and rules we have; these differ in different systems, for example natural deduction or a Hilbert style axiom system, but it is easy in all cases) based on the idea that if something holds for all x and y, it also holds for all x and y such that y=|x|, or y=x, as pointed out in an earlier post.

But of course, I can try to convince you with the semantic method that ##\forall a[a\neq 0\Longrightarrow a^2>0]## is true.

Case 1: ##a=0##. Then ##a\neq 0## is false and ##a^2> 0## is false. Hence, the implication ##a\neq 0\Longrightarrow a^2> 0## is true.

Case 2: ##a>0##. Then ##a\neq 0## is true and ##a^2> 0## is true. Hence, the implication ##a\neq 0\Longrightarrow a^2> 0## is true.

Case 3: ##a<0##. Then ##a\neq 0## is true and ##a^2> 0## is true. Hence, the implication ##a\neq 0\Longrightarrow a^2> 0## is true.

Thus, ##a\neq 0\Longrightarrow a^2> 0## is true for all (real) values of ##a##. In other words, ##\forall a[a\neq 0\Longrightarrow a^2>0]## is true.

 Quote by Erland If you want to prove that something is a theorem, you should make a derivation from the axioms. My intention with the "semantic" method was not to prove that the sentence in question is a theorem, but to convince you that it is true, so that the proposed definition of absolute value does not lead to something false. The sentence in the previous post can be derived in a few steps from the given definition of absolute value, by using logical axioms and rules of inference (exactly how depends on which logical axioms and rules we have; these differ in different systems, for example natural deduction or a Hilbert style axiom system, but it is easy in all cases) based on the idea that if something holds for all x and y, it also holds for all x and y such that y=|x|, or y=x, as pointed out in an earlier post. But of course, I can try to convince you with the semantic method that ##\forall a[a\neq 0\Longrightarrow a^2>0]## is true. Case 1: ##a=0##. Then ##a\neq 0## is false and ##a^2> 0## is false. Hence, the implication ##a\neq 0\Longrightarrow a^2> 0## is true. Case 2: ##a>0##. Then ##a\neq 0## is true and ##a^2> 0## is true. Hence, the implication ##a\neq 0\Longrightarrow a^2> 0## is true. Case 3: ##a<0##. Then ##a\neq 0## is true and ##a^2> 0## is true. Hence, the implication ##a\neq 0\Longrightarrow a^2> 0## is true. Thus, ##a\neq 0\Longrightarrow a^2> 0## is true for all (real) values of ##a##. In other words, ##\forall a[a\neq 0\Longrightarrow a^2>0]## is true.
Let us take your proof step by step.

How do you know that ##a^2>0##
is false if a =0

 Quote by stauros Let us take your proof step by step. How do you know that ##a^2>0## is false if a =0
I repeat: If you want a PROOF, make a formal derivation from the axioms! The intention of the "semantical" method was NOT to do that, but to give arguments that, hopefully, would convince you. This seems to have failed. If you are not already convinced that ##0^2>0## is false, nothing else I say will convince you of that, except possibly a derivation from the axioms.

But what would be the point of that, since you anyway question simple consequences of the axioms and inference rules of logic (which differ in different systems but have the same power to prove theorems), such that one can go from ##\forall x\forall y P(x,y)## to ##\forall x P(x,x)##?
This latter can be derived in few steps with any logical system in use. As an exercise, derive with the system in your logic textbook (and I don't know which book you have, so I cannot do it until you tell me which system is used in it).