Formalization of sqroot definition

[peos69]

2) Which definition to use?

.

Your definition of course

 

[CompuChip]

CONGRATULATIONS YOU DECIDED to learn the famous step wise semi formal proof, i wonder who was the inventor of that proof.

You mean who was the one that gave it first? I don't think for such trivial theorems this is relevant...

NEARLY EVERY h.school textbook and indeed in the lower classes proves that theorem,would you care to look,there it says for all x.

In logic, "for all x" means: "for all x in the universe of discourse." If the square root is only defined for positive x, you can only prove things about it for positive x.

But let us see if the theorem holds for x<0.
let x=-3 then sq root(-3)^2=sq root(9)=3.
Also abs value(-3)=3
Hence sq root(x)^2= abs value(x)

You should be clear about your notation.
( \sqrt{-3} )^2 = -3 \stackrel{!}{\neq} \sqrt{(-3)^2} = 3.
This is even more easy to prove, as: (Let x > 0):
(-x)^2 = x^2 so \sqrt{(-x)^2} = \sqrt{x^2} = x (I'm "guessing" a root here and by injectivity properties it is the unique one). And by the way absolute value works, x = |-x|.

IF YOU CANNOT prove a theorem it does not mean it is not correct

Who said otherwise?

SO NOW YOU found out that your definition WAS INCOMPLETE.
I wonder WHAT NEXT

Actually my definition was perfectly correct. Especially since you keep stressing "high school"... I don't know about you but I didn't learn about complex numbers until after high school, so the definition on \mathbb{R}_+ is perfectly all right. Also the proof I gave was perfectly correct and the theorem was true as well. But then you started about x < 0, so I extended (not: completely changed!) the definition and showed that the theorem was wrong and proved the correct version of it.

OF COURSE your definition of sq root is pathetic let alone wrong. [...] boring,thats a word used by those that they cannot do it
IN TOUGH analysis where a lot of theorems are incomplete ,wrong or at least badly written
[...]

I think I'm done in this thread, I'll see you when you've decided to do real mathematics.

 

[uman]

Theorem. The set of all nonnegative real numbers whose square is less than some positive real number a has a supremum.
Proof.
If a>=1, x^2 < a implies x < a < a+1, since x^2<1 if x<1 and x < x^2 if x > 1.
If a<1, x^2 < a implies x < 1 < a+1. In either case, the set in question is bounded above by a+1. Since 0 is in the set, it is nonempty and therefore has a supremum by the least upper bound axiom.

Notation. Let R* be the set of nonnegative real numbers.

Definition. Let the "square root function" be the relation f: R* -> R* defined as follows:
1) (0,0) is in f.
2) If x > 0, let y denote the supremum of the set of nonnegative real numbers whose square is less than x. Then (x,y) is in f.
3) No other ordered pair is in f.

Theorem. f is a function.
Proof.
If (0,a) is in f, 1), 2), and 3) together show that a must equal 0.
If (x,a) is in f and x != 0, 3) and 2), as well as the fact that the supremum of a set of real numbers is unique when it exists, show that a is uniquely determined by x.
If x is a real number greater than or equal to zero, (x,a) is in f for some a.
The first two statements show that f is defined everywhere on its domain, and the last shows that the value of f(x) is uniquely determined. Hence x is a function.

Theorem. f(x^2) = abs(x).
Proof. Assume x is positive. Then f(x^2) is the supremum of the set of real numbers y such that y^2 < x^2. If f(x^2) > x, let a = (f(x^2) + x)/2 < f(x^2). Since f(x^2) is the supremum of the set of real numbers y such that y^2 < x^2, we should have a^2 < x^2. But a > x, so a^2 > x^2. It is similarly easy to prove that f(x^2) is not less than x, so f(x^2) = x = abs(x).

Assume x is negative. Then (-x) is positive, and (-x)^2 = x^2. We may apply the statement just proven to obtain f(x^2) = f((-x)^2) = abs(-x) = abs(x).

f(0) = 0 = abs(0), by the definition of f.

These three statements together show that f(x^2) = abs(x).

 

[peos69]

If somebody asks you to teach him how to drive a car you put in an airspace bus and you send him to the moon, or if a kid ask you to teach him how to count you do it thru
set theory??

 

[peos69]

Sorry i should have said ,wrong set theory instead of only set theory

 

[CompuChip]

Peos you seem to be convinced that you are right and all of us are wrong. Unless you can clearly explain your problem instead of repeating that our mathematics is not right, I don't think anyone can help you.

 

[Gokul43201]

Unless you can clearly explain your problem ...

After all this, you want to give peos another chance? What if you find out that peos made another unacknowledged notation error 3 pages too late? Why waste your time?

 

[arildno]

What is the problem here?
Let us take the high-school "definition" of the sqrt of a non-negative number "a" as the non-negative number that multiplied with itself equals a, i.e:
\sqrt{a}*\sqrt{a}=a
Or, equivalently:
(\sqrt{a})^{2}=a
As for proving \sqrt{a^{2}}=|a|
we note that \sqrt{a^{2}} must satisfy the equation:
x^{2}=a^{2}
From this equation, it follows that either x=a, OR, x=-a.
But in our defintion of the square root, we required that it has to be the NON-negative number with that property!
Thus, it follows that x=a IF a>=0, and x=-a IF a<0.
But this is precisely what is encapsulated in the concept of the "absolute value" of a number, hence we have proven that:
\sqrt{a^{2}}=|a|

 

[peos69]

Assume x is negative. Then (-x) is positive, and (-x)^2 = x^2. We may apply the statement just proven to obtain f(x^2) = f((-x)^2) = abs(-x) = abs(x).

f(0) = 0 = abs(0), by the definition of f.

These three statements together show that f(x^2) = abs(x).

WHEN X<0 THEN abs(x)=-x AND NOT abs(x) = x
LEARN THE DEFINITION of the absolute value of a REAL No

 

[CompuChip]

What is the problem here?

Peos69. Otherwise, none.

 

[CRGreathouse]

Assume x is negative. Then (-x) is positive, and (-x)^2 = x^2. We may apply the statement just proven to obtain f(x^2) = f((-x)^2) = abs(-x) = abs(x).

f(0) = 0 = abs(0), by the definition of f.

These three statements together show that f(x^2) = abs(x).

WHEN X<0 THEN abs(x)=-x AND NOT abs(x) = x
LEARN THE DEFINITION of the absolute value of a REAL No

uman didn't claim that abs(x) = x for x negative, just that abs(-x) = abs(x).

 

[peos69]

arildo your proof is right but where is the formalization of the definition??
Now we started getting somewhere

 

[peos69]

uman didn't claim that abs(x) = x for x negative, just that abs(-x) = abs(x).

HE DID the fatal mistake to use a formula which he proved ONLY FOR X>=0
definitely his formula does not apply for x<0,unless he proves (assuming he proves),
that f(x^2)= -x

 

[peos69]

After all this, you want to give peos another chance? What if you find out that peos made another unacknowledged notation error 3 pages too late? Why waste your time?

HOW MANY have i done up to now??
NAME them

 

[peos69]

Peos you seem to be convinced that you are right and all of us are wrong. Unless you can clearly explain your problem instead of repeating that our mathematics is not right, I don't think anyone can help you.

Definitely i have no problem the problem is yours that you cannot give a proper formalization of a h.school concept
You must also remember i did not discovered the sqroot

 

[arildno]

arildo your proof is right but where is the formalization of the definition??
Now we started getting somewhere

What nonsense are you talking?
My definition is perfectly formalized, in that particular format known as plain, English language.

 

[Gokul43201]

HOW MANY have i done up to now??
NAME them

Try these for size:

1)for all x sqoot (x)^2 = absolute value (x)

And several posts later...

let x=-3 then sq root(-3)^2=sq root(9)=3.

Both have parentheses in the wrong place. This was pointed out more than once, and you've chosen not to acknowledge it.

And quit with the capitals already, if you don't wish to get booted from here.

 

[peos69]

What nonsense are you talking?
My definition is perfectly formalized, in that particular format known as plain, English language.

FORMALIZATION IS FREE from any language that's the beauty of it

 

[arildno]

FORMALIZATION IS FREE from any language that's the beauty of it

Nonsense.
Besides, I note that you didn't bother to heed Gokul's WARNING.

 

[HallsofIvy]

I suspect that one problem is that after 68 posts in this thread you have never said exactly what you mean by "formalization".

 

[peos69]

:



, if you don't wish to get booted from here.

nice expression,good english

 

[uman]

peos69: Which part of my post only applies to x >= 0? The definition of f? That's true, but in high school I don't think most students learn about the square root of negative numbers; that is, they consider the square root function as a real-valued function. It would not be too difficult to extend my definition and proofs to make f complex-valued, and defined everywhere on R.

are you saying that I only proved that f(x^2) = abs(x) if x>=0? You are mistaken. I proved first that f(x^2) = abs(x) if x>0, then I used that fact to prove that f(x^2) = abs(x) if x<0. Then I pointed out that f(0) = 0. So,
If x > 0 f(x^2) = |x|
if x = 0 f(x^2) = |x|
if x < 0 f(x^2) = |x|. So in all cases, f(x^2) = |x|.
Do you see any specific problems with my proof? Or any other problems with the theory of the square root function that I developed in my post? I wrote it at about three in the morning and may well have made some mistakes!

Also, what precisely do you mean by "formalise"? I think if you made that clear we would be able to help you much better.

Please answer ALL the questions in this thread before posting again.

 

[berkeman]

nice expression,good english

peos69, Gokul is pointing out that we have Rules here on the PF, and it all works best if we check our attitudes at the door, and discuss the math and physics in a straightforward way. The Mentors are monitoring this thread, obviously.

 

[peos69]

Is it in your rules to use expressions "if you don't wish to get booted from here"
I am surprised.

 

[berkeman]

Is it in your rules to use expressions "if you don't wish to get booted from here"
I am surprised.

It is a common expression in English, and not considered rude in this context. Just keep a calm attitude, and focus on the math and the physics. And as Gokul was pointing out, all capital letters is considered yelling in the context of forum discussions, and should be avoided.

 

[peos69]

can i then in a common expression in English say to him boot yourself?
capital letters are considered for emphasizing things

 

[CompuChip]

You can emphasize things using italics with the [ i] [/ i] tag (without spaces), for example. That is already a lot nicer (and using Control + I, no more work than pressing Caps Lock twice).

the problem is yours that you cannot give a proper formalization of a h.school concept

The problem is not mine, because I am perfectly happy with my "formalization". If you are not, you should specify what you mean by a "formalization" (as pointed out repeatedly) and what there is as a "concept" to the square root.

 

[berkeman]

can i then in a common expression in English say to him boot yourself?

No, at this point that would be rude. It has nothing to do with the math and physics of this thread. He was pointing out the fact that you could receive infraction points for the aggressive nature of your posts. You would not be pointing out anything useful with the reply that you propose.

capital letters are considered for emphasizing things

Not in the context of web forum discussions, especially not the way you used them.

This thread appears to have run its course. Thread locked.

 

[HallsofIvy]

Note that "boot yourself" is quite different in meaning from "booted from here". The first implies physical violence while the second doesn't.

What has happened so far is that many people have given different types of "formalization" of the square root function and you have rejected them all. Will you please tell us what you mean by "formalization"?

 

[berkeman]

Will you please tell us what you mean by "formalization"?

Actually, that might be kind of hard for him. I already locked the thread. It's probably best to just let it lie, unless you want to unlock and carry on. I'm fine either way.