for the following problem:

 

For all ,a : if [math]0\leq a\leq 1[/math], then [math]\sqrt{1+\sqrt{1-a^2}} +\sqrt{1-\sqrt{1-a^2}} =\sqrt{2+2a}[/math].....................................................................................A

 

the following proof is suggested:

 

[math]0\leq a\leq 1\Longrightarrow 0\leq 1-a^2\leq 1\Longleftrightarrow 0\leq \sqrt{1-a^2}\leq 1[/math][math]\Longrightarrow(\sqrt{1+\sqrt{1-a^2}}\geq 0\wedge \sqrt{1-\sqrt{1-a^2}}\geq 0)\Longrightarrow[/math] [math](\sqrt{1+\sqrt{1-a^2}}+\sqrt{1-\sqrt{1-a^2}})^2[/math] = [math]1+\sqrt{1-a^2}+1-\sqrt{1-a^2}+2\sqrt{(1+\sqrt{1-a^2})(1-\sqrt{1-a^2})}=2+2|a|=2+2a[/math]

 

And since [math] 2+2a\geq 0[/math] (A) is valid

That derivation is fine. What is your objection?

That derivation is fine. What is your objection?

I was wandering what are the theorems or definitions used in the proof

Isn't it more than a bit self-evident in this case?

 

Some kindly advice: If you wish to progress in your mathematics education (self-education?) it would behoove to drop this obsessive-compulsive behavior of yours regarding formality. It is heeding your progress.

There is always some question of rigour in any proof. It can be a question of where you start and what theorems/propositions/corollaries etc you accept as being proved. Not to mention what you would expect your reader to accept.

 

However, in the examples you have given us you should take the basic properties of real numbers of being accepted.

 

I suppose you could go back to an axiomatic construction of the real numbers (something like Tarski's) but how would this really help you?

 

It is good that you are thinking about these things, but be cautious you do not hinder yourself we unnecessary formality.

I propose for the above proof the following set of theorems:

 

1) for all ,x,y : ( [math]x\geq 0[/math] and [math]y\geq 0\Longrightarrow ( x\leq y\Longleftrightarrow x^2\leq y^2)) [/math] or in words:

 

for all ,x,y : if x and y are greater or equal to zero ,then x is lees or equal to y iff x to the square is less or equal to y to the square.

 

2) for all ,x,y,z : [math]x\leq y\leq z\Longleftrightarrow -x\geq -y\geq -z[/math]

 

3) for all ,x,y,z,w :[math] x\leq y\leq z\Longleftrightarrow x+w\leq y+w\leq z+w[/math]

 

4) for all ,x,y : [math] x\geq 0[/math] and [math]y\geq 0\Longrightarrow( x\leq y\Longleftrightarrow\sqrt x\leq\sqrt y)[/math]

 

5) for all x : [math] x\geq 0\Longrightarrow \sqrt x\geq 0[/math]

 

6) for all x : [math] x\geq 0\Longrightarrow(\sqrt x)^2 = x[/math]

 

7) [math] 1\geq 0[/math]

 

8) for all x,y : [math]x\geq 0[/math] and [math]y\geq 0\Longrightarrow\sqrt x.\sqrt y = \sqrt xy[/math]

 

9)for all ,x,y : [math] ( x+y)^2 = a^2 +b^2+2ab[/math]

 

10) for all ,x,y : [math] (x+y)(x-y) = x^2-y^2[/math]

 

And finally the definition of sqrt.

 

Can any one propose a different or shorter list

Quite seriously ajb and DH know what they are talking about and it would be a good idea to y'know, listen to him.

 

But while we're here:

 

9 and 10 are both special cases of [imath]x(a+b)=xa+xb[/imath], which is one of the field axoims anyway, so they are clearly true (which means that you wouldn't bother going through it rigoursly unless you were made to for the sake of an exercise).

 

6 is nearly how you'd define sqrt in the first place.

 

4 and 5 are made redudant by 1.

 

1,2,3,7 and 8 might be worth proving once but really it doesn't matter if you don't.

Have a look at the synthetic approach to real numbers.

 

On Wiki.

On PlanetMath.

 

I expect, with some work you could prove all you need from these axioms.

On this note. I was thinking today proving that a positive real number has exactly one positive real square root might not be so trivial. The only proof that I could think of was dependent on ideas from real analysis.

Quite seriously ajb and DH know what they are talking about and it would be a good idea to y'know, listen to him.

 

But while we're here:

 

9 and 10 are both special cases of [imath]x(a+b)=xa+xb[/imath], which is one of the field axoims anyway, so they are clearly true (which means that you wouldn't bother going through it rigoursly unless you were made to for the sake of an exercise).

 

6 is nearly how you'd define sqrt in the first place.

 

4 and 5 are made redudant by 1.

 

1,2,3,7 and 8 might be worth proving once but really it doesn't matter if you don't.

.

 

In which way 4 and 5 are redundant by 1

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Have a look at the synthetic approach to real numbers.

 

On Wiki.

On PlanetMath.

 

I expect, with some work you could prove all you need from these axioms.

 

In taking the first implication of the proof i.e

 

[math] 0\leq a\leq 1\Longrightarrow 0\leq 1-a^2\leq 1[/math]

 

what would be the theorems that are involved and how are they involved

On this note. I was thinking today proving that a positive real number has exactly one positive real square root might not be so trivial. The only proof that I could think of was dependent on ideas from real analysis.

Assume [math]c>0[/math] has two distinct positive real roots, call them [math]a[/math] and [math]b[/math]: [math]a^2=b^2=c[/math]. Since these roots are distinct, one will be smaller than the other. Denote this lesser root [math]a[/math]: [math]0<a<b[/math]. Multiplying both sides of an inequality by the same positive number does not change the nature of the inequality. Multiplying both sides of [math]a<b[/math] by [math]a[/math] yields [math]a^2<ab[/math] while multiplying both sides by [math]b[/math] yields [math]ab<b^2[/math]. As less than is a transitive relationship, [math]a^2<b^2[/math], and this contradicts the initial assumption that [math]a^2=b^2[/math].

triclino:

For 4, just take 1 and set X=sqrt(x) and Y=sqrt(y). Similarly for 5, set X=sqrt(x) and Y=0. All you've done is give specific values to the variables, so it is the same lemma.

 

D H:

That's sufficient to prove that a positive real has no more than one positive root. Proving that it has at least one would in fact be much easier: I think I just ran into a brick wall by trying to do both at once, thanks though.

triclino:

For 4, just take 1 and set X=sqrt(x) and Y=sqrt(y). Similarly for 5, set X=sqrt(x) and Y=0. All you've done is give specific values to the variables, so it is the same lemma.

 

D H:

That's sufficient to prove that a positive real has no more than one positive root. Proving that it has at least one would in fact be much easier: I think I just ran into a brick wall by trying to do both at once, thanks though.

 

For (5): if you put x=sqrt(x) and y=0 in (1) you have:

 

[math]\sqrt x\geq 0[/math] and [math]0\geq 0\Longrightarrow (\sqrt x\leq 0\Longleftrightarrow (\sqrt x)^2\leq 0)[/math]

 

Where you have : [math] \sqrt x\geq 0[/math] and [math]\sqrt x\leq 0[/math] and you cannot get (5) in any way.

 

As for formula (4) it is equivalent to (1) ,hence not redundant

Oh, x's and y's the other way around, whatever.

 

Equivalence does sort of carry redundancy with it, when the equivalence is trivial.

Oh, x's and y's the other way around, whatever.

 

Equivalence does sort of carry redundancy with it, when the equivalence is trivial.

 

If you put x=0 and y =sqrt(y) we have:

 

 

[math] 0\geq 0[/math] and [math]\sqrt y\geq 0\Longrightarrow(0\leq\sqrt y\Longleftrightarrow 0\leq(\sqrt y)^2)[/math]

 

From which formula you can never prove :

 

[math]y\geq 0\Longrightarrow\sqrt y\geq 0[/math]

 

unless you assume:

 

1) [math] y\geq 0\Longrightarrow\sqrt y\geq 0[/math],a formula that you want to prove ,and:

 

2) [math] y\geq 0\Longrightarrow(\sqrt y)^2 = y[/math]

 

 

Also in proving the equivalence between (1) and (4) it is not a trivial matter because we have other theorems involved in the proof and the proof it is not just a substitution

Obviously:

 

[math]\sqrt{y}[/math] is [math]\geq 0[/math]

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Unless y is 0, sqrt(y) is greater than 0. Tell me a number like that which is less than zero. Negative numbers? You'd get negative numbers. Is Y a composite? A positive?

 

Whatever it is; I think listening to D H and ajb is the best way to go.

Obviously: [math]\sqrt{y}[/math] is [math]\geq 0[/math]

We're talking axiomatics here, 'obvious' just doesn't cut it. The SQRT operator is usually taken to mean the positive square root but only by convention since any number with a strictly positive square root also has a negative one.