What Mathematical Principle Justifies the Equation x=a+b+c?

[evagelos]

what allow us to put x=a+b+c

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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??

 

[Redbelly98]

Can you post (or link to) a specific example of where you've seen this?

 

[evagelos]

inequalities systems of equations e.t.c e.t.c for example one that comes into my mind just now is the following inequality:

(a+b+c)/3 >= (abc)^1/3 here one way of solving the inequality is to use the substitution

x=(a+b+c)/3 of course without using Am-Gm

 

[Redbelly98]

Hmmm ... without knowing how to solve that one myself, it seems that this is a way of simplifying things by reducing the number of variables in the problem.

What allows us to do this is the fact that, in this particular problem, "x" had not been defined as anything up to that point. So we are free to define it as (a+b+c)/3, or as anything else we wish.

Hope that's clear enough ... perhaps others in here are familiar with this example and can elaborate a little more than I have?

 

[evagelos]

I am sorry for the example it was brought up at random ,its proof is immaterial

i can give other examples

 

[evagelos]

.What allows us to do this is the fact that, in this particular problem, "x" had not been defined as anything up to that point.

can this be considered as axiom or a theorem.

because the above is your basic argument in allowing the substitution

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

 

[Redbelly98]

What allows us to do this is the fact that, in this particular problem, "x" had not been defined as anything up to that point.

can this be considered as axiom or a theorem.

because the above is your basic argument in allowing the substitution

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

Honestly I'm not sure if that's considered an axiom or theorem ... or neither.

This is done all the time in word problems, such as:

Jim is twice as old as Jane was 4 years ago, and three times as old as she was 6 years ago. How old are Jim and Jane?

And to find the answer, we might begin by saying:
"Let x = Jim's age now, and y = Jane's age now ... "

 

[cristo]

This is just basic algebra: we let symbols represent quantities. I'm not sure why this would cause confusion, though.

 

[Dragonfall]

It comes from the axiom of the magic bag of variables.

 

[evagelos]

Honestly I'm not sure if that's considered an axiom or theorem ... or neither.

This is done all the time in word problems, such as:

Jim is twice as old as Jane was 4 years ago, and three times as old as she was 6 years ago. How old are Jim and Jane?

And to find the answer, we might begin by saying:
"Let x = Jim's age now, and y = Jane's age now ... "

i think this is mathematical modeling isn't it

 

[HallsofIvy]

Yes, any application of mathematics is mathematical modeling.

And we are free to define x to be a+ b, as long as we haven't already used "x" for some other meaning because we can always just assign a name to something.

 

[evagelos]

now i suppose you pushing me to open a new thread or to ask here what is the definition of the definition,because we all might not agree what a definition is.
Because when you say we define x=a+b i don't think define is the proper word...
for me

 

[Redbelly98]

It might help if you said just where you are in your studies of math ... clearly you've already had some algebra, and I'm guessing geometry also (where theorems and axioms are introduced, at least in many USA high school curricula). Perhaps you are well beyond even these, and just looking for formal justification of what many take for granted.

I checked a h.s. geometry book I have. It says we have four basic "rules", or types of justification, that allow us to proceed from step to step in a proof or derivation:

Axioms
Theorems
Definitions
Given information

If we have only these 4 choices, I think "definition" comes closest to what you are looking for. If there is anything else outside of these, I am not aware of it.

 

[HallsofIvy]

now i suppose you pushing me to open a new thread or to ask here what is the definition of the definition,because we all might not agree what a definition is.
Because when you say we define x=a+b i don't think define is the proper word...
for me

Then I recommend you get a dictionary and look up the definition of "definition"!

 

[snipez90]

inequalities systems of equations e.t.c e.t.c for example one that comes into my mind just now is the following inequality:

(a+b+c)/3 >= (abc)^1/3 here one way of solving the inequality is to use the substitution

x=(a+b+c)/3 of course without using Am-Gm

I'm curious as to see how you would prove AM-GM for three variables using that substitution. I know you can prove it by factoring $$a^3 + b^3 + c^3 -3abc$$ but it's not obvious to me how such a substitution helps.

 

[evagelos]

as i said in post #5 the example was picked up at random from a book the proof is not mine .The book uses the Am-Gm for two variables plus the substitution in solving
the inequality

 

[snipez90]

But you mentioned that it doesn't use AM-GM and now you're saying it does involve AM-GM of two variables. Anyways that doesn't matter. I'm sure you can post the proof? I'm just curious that's all.

 

[evagelos]

But you mentioned that it doesn't use AM-GM and now you're saying it does involve AM-GM of two variables. Anyways that doesn't matter. I'm sure you can post the proof? I'm just curious that's all.

here is a proof direct from the book:

let x=(a+b+c)/3 = (a+b+c+x)/4 = [(a+b)/2 +(c+x)/2]/2 >= [sqroot(ab)+sqroot(cx)]/2>=

sqroot[sqroot(ab)sqroot(cx)]= sqroot[sqroot(abcx)] = (abcx)^1/4

hence x=(abcx)^1/4 ====> x^4>=abcx

if x=0 then a=b=c=o.
if x>0 then : x^3>= abc<====> x>= (abc)^1/3 <====> (a+b+c)/3>= (abc)^1/3

 

[snipez90]

That is a tricky substitution. Very nice proof, thanks!

 

[evagelos]

as i said the proof is not mine

 

[evagelos]

And we are free to define x to be a+ b, as long as we haven't already used "x" for some other meaning because we can always just assign a name to something.

you cannot define a concept without 1st proving its existence.For example you cannot define sqroot unless you prove that every No has a sqroot,so before you define x=a+b,you must prove 1st that there exists such an x

 

[Focus]

you cannot define a concept without 1st proving its existence.For example you cannot define sqroot unless you prove that every No has a sqroot,so before you define x=a+b,you must prove 1st that there exists such an x

The x exists for sure, I think what you mean is you must prove that x is an element of a group or a set. Square root is also defined on any number, but if you square root a real number it might not lie in the reals.

 

[evagelos]

you cannot define a concept without 1st proving its existence.For example you cannot define sqroot unless you prove that every No has a sqroot,so before you define x=a+b,you must prove 1st that there exists such an x

The x exists for sure, I think what you mean is you must prove that x is an element of a group or a set. Square root is also defined on any number, but if you square root a real number it might not lie in the reals.

the positive root of the equation , x^2=y , is defined as the sqroot of y denoted by sqroot(y) .

the question now is if we give any values to y will the above equation have a +ve asqroot ??

the following theorem answers that question:

,,,,,,,,,,,,,,,,,,,(y)[ y>=0-------->E!x( x>=0 & x^2=y)].........,,1

So that theorem guarantees the existence of sqroot(y).........

and then we define the sqroot concept ,otherwise we might have a Vacuum definition.

In the same way we must prove the existence of x and then define it

 

[Focus]

the positive root of the equation , x^2=y , is defined as the sqroot of y denoted by sqroot(y) .

the question now is if we give any values to y will the above equation have a +ve asqroot ??

the following theorem answers that question:

,,,,,,,,,,,,,,,,,,,(y)[ y>=0-------->E!x( x>=0 & x^2=y)].........,,1

So that theorem guarantees the existence of sqroot(y).........

and then we define the sqroot concept ,otherwise we might have a Vacuum definition.

In the same way we must prove the existence of x and then define it

The theorem does not answer the question. You have not stated what x is. For all I know x could be a doughnut, are you trying to square root a doughnut. When you say something exists, you need a set in which it exists. Its utterly useless to say it just exists without specififying in which set or group.

 

[evagelos]

When we put x=a+b or x=a+b+c ,where are we ,in a set with doughnuts?

The theorem does not answer the question ,it simply shows you that before we define a
concept we must have an existence theorem , as it happens with the sqroot in the set
of real Nos