Many times in mathematics in proving something we can put

x=a+b+c or x= a-b^2 or x=y e.t.c.

What axiom or theorem in mathematics allow us to do that??

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- Aug 21st 2008, 02:06 PM #1

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- Aug 21st 2008, 06:43 PM #2

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None. We are just using what the problem is saying. Say that $\displaystyle f: \mathbb{R}\to \mathbb{R}$ is such a function so that $\displaystyle f(x) = f(\tfrac{x}{2})$. This means $\displaystyle f(2y) = f(y)$ for any $\displaystyle y\in \mathbb{R}$. Thus, $\displaystyle f$ has property that $\displaystyle f(x) = f(2x)$ for any $\displaystyle x$.

This is just using the definition of what the function is.

- Aug 22nd 2008, 06:50 AM #3

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Very interesting your function concept,but then again what allow us for the creation of a function.

in the following example how the function concept could be used?

suppose we want to prove that:

...................(a+b+c)/3 >= (abc)^1/3,with a,b,c>=0

Now without using the Am-Gm concept for the proof of the inequality,

one way of solving the inequality is to use the substitution

......................x=(a+b+c)/3

- Aug 22nd 2008, 10:13 AM #4

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- Sep 5th 2008, 06:25 PM #5

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- Sep 6th 2008, 05:34 PM #6

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- Sep 8th 2008, 12:01 PM #7
I would like to give a nice proof for this inequality:

$\displaystyle (abc)^{1/3}$

$\displaystyle =\exp\big\{\ln\big((abc)^{1/3}\big)\big\}$

$\displaystyle =\exp\bigg\{\frac{1}{3}\ln(a)+\frac{1}{3}\ln(b)+\f rac{1}{3}\ln(c)\bigg\}$

$\displaystyle \leq\frac{1}{3}\exp\big\{\ln(a)\big\}+\frac{1}{3}\ exp\big\{\ln(b)\big\}+\frac{1}{3}\exp\big\{\ln(c)\ big\}\text{ convex function property}$

$\displaystyle =\frac{1}{3}\big(a+b+c\big),$

which is the desired result (also see the proof for Young's inequality).

As ThePerfectHacker mentioned, there is no axiom on the things you told.

You just find your way yourself by picking new parameters or arranging the given ones to reveal your own way.

Ofcourse, experience is the most important thing in this dark forest!

- Sep 8th 2008, 04:54 PM #8

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In my very 1st post i asked:

**......what allows us to put x=a+b+c ...............................................**

in a proof that uses this substitution.And then i brought up as an**example**to the posts of the PerfectHacker the said inequality

Anyway in your proof you do not use ,if i am not mistaken,any substitutions

PerfectHacker brought in the concept of function and i was forced to point out that even the concept of function needs a theorem of existence.

So now we are faced with two questions ;

1)If the concept of function needs a theorem of existence,or

2)The substitution**x=a+b+c, or any substitution of this type**need a theorem or axiom to back it up

For example the concept of sqroot need a theorem of existence before it is defined

Notice, my ' or" is not exclusively disjunctive

- Sep 8th 2008, 05:02 PM #9

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- Sep 8th 2008, 05:05 PM #10

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Are you asking why we know there is such a function $\displaystyle f:\mathbb{R}\to \mathbb{R}$ such that $\displaystyle f(x) = f(\tfrac{x}{2})$? Is that why you mean by existence? If so it does not matter - that was simply for an example. We can assume it exists. And see where we get from there.

- Sep 8th 2008, 05:18 PM #11

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- Sep 8th 2008, 05:25 PM #12

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Yes you can assume. In math we always make assumptions.

However, if a proof depends on the assumption it is not a proof. You need to prove the assumption.

In the example I gave you there is nothing bad. I was not trying to prove anything. Just give you an example. Everything was justified.