Is There a Real Solution for a^n+b^n=c^n as n Approaches Infinity?

[Icebreaker]

I'm trying to show the following:

Let $$a^n+b^n=c^n$$ where $$a, b, c$$ are not equal to zero. There is no real solution for $$a, b, c$$ as $$n\rightarrow\infty$$.

Proof:

Assume the contrary. If there is a real solution for $$a, b, c$$ as $$n\rightarrow\infty$$, then

$$lim_{n\rightarrow\infty} \frac{a^n+b^n}{c^n}=1$$

must also be true.

We can use properties of limits and algebra to obtain, from the previous equation

$$lim_{n\rightarrow\infty} \frac{a^n}{c^n}+lim_{n\rightarrow\infty} \frac{a^n}{c^n}=1$$

It can be shown that if $$a^n+b^n=c^n$$ (where $$a, b, c$$ are not equal to zero) is true, then $$c>a$$ and $$c>b$$

Therefore

$$0<\frac{a}{c}<1$$ and $$0<\frac{b}{c}<1$$

Using properties of limits, we can state that

$$lim_{n\rightarrow\infty} \frac{a^n}{c^n} = lim_{n\rightarrow\infty} \frac{b^n}{c^n} = 0$$

and therefore

$$lim_{n\rightarrow\infty} \frac{a^n}{c^n}+lim_{n\rightarrow\infty} \frac{a^n}{c^n}=lim_{n\rightarrow\infty} \frac{a^n+b^n}{c^n}=0$$

Which is not 1, and we have thus shown that the contrary to the first statement cannot be true.

p.s. I hate tex

 

[robert Ihnot]

Icebreaker,
$$lim_{n\rightarrow\infty} \frac{a^n}{c^n}+lim_{n\rightarrow\infty} \frac{a^n}{c^n}=1$$

for starters where did you get this?

Before we get too caught up in that, I have to ask, What does a limit have to do with it?

 

[Icebreaker]

$$lim_{n\rightarrow\infty}\frac{a^n+b^n}{c^n} =lim_{n\rightarrow\infty}\frac{a^n}{c^n}+\frac{b^n}{c^n} =lim_{n\rightarrow\infty} \frac{a^n}{c^n}+lim_{n\rightarrow\infty} \frac{a^n}{c^n} =1$$

If you have a way to prove it without limits, be my guest.

 

[Curious3141]

First of all, I don't like the problem "statement" at all, it makes little mathematical sense. Did you get that out of a textbook, or did you make it up ?

The ("Fermat-like") equation $$a^n + b^n = c^n$$ has infinitely many real solution sets (a,b,c) for every positive integral value of n. Asking what solutions exist when n is infinite, or even "n tends to infinity" is meaningless. This is not evaluating the limit of an expression, this is gobbledegook.

 

[Icebreaker]

Why is asking what solutions exist when n increases without bound meaningless?

 

[Curious3141]

Think about what you're asking.

Do real solution sets (a,b,c) exist for n = 1000 ? YES

Do real solution sets (a,b,c) exist for n = 100,000 ? YES

Do real solution sets (a,b,c) exist for n = 10^(10^(10^10) ? YES

Is n "tending" to infinity in the above ? I suppose so.

Do real solutions exist all the way regardless of the finite value of n, no matter how large ? Obviously.

Does it make sense to ask the question for an infinite value of n ? NO.

***

Do you see now ?

 

[Icebreaker]

Explain what mathematical principal I'm violating here? Because from what I see above I have every right asking the question I did.

 

[Curious3141]

Explain what mathematical principal I'm violating here? Because from what I see above I have every right asking the question I did.

As has been discussed a very large number (but still finite) number of times in this forum, infinity is NOT a number. You cannot set up an equation using infinity and expect it to have a solution.

If, OTOH, you're asking whether that equation has solutions for very large finite values of n, then the answer is always YES, so the premise you want to prove is wrong.

When you are evaluating the limits of expressions as something "tends" to infinity, all you are doing is investigating the behaviour or a mathematical expression as one of the terms is made arbitrarily large. Note that even if the limit exists, the expression can NEVER actually equal the limit. It can only approach it.

Just as you cannot speak of the expression ever having the ACTUAL value of the limit, you cannot set up an equation using those rules.

 

[Icebreaker]

As has been discussed a very large number (but still finite) number of times in this forum, infinity is NOT a number. You cannot set up an equation using infinity and expect it to have a solution.

Oh? $$lim_{n\rightarrow\infty}\frac{1}{n}=0$$

If, OTOH, you're asking whether that equation has solutions for very large finite values of n, then the answer is always YES, so the premise you want to prove is wrong.

Of course there's a solution for every finite n. So?

When you are evaluating the limits of expressions as something "tends" to infinity, all you are doing is investigating the behaviour or a mathematical expression as one of the terms is made arbitrarily large.

So I'm investigating the value of the expression as a variable is increasing without bounds. And?

Just as you cannot speak of the expression ever having the ACTUAL value of the limit, you cannot set up an equation using those rules.

I never said n EQUALS infinity.

 

[Icebreaker]

Think about what you're asking.

Do real solution sets (a,b,c) exist for n = 1000 ? YES

Do real solution sets (a,b,c) exist for n = 100,000 ? YES

Do real solution sets (a,b,c) exist for n = 10^(10^(10^10) ? YES

Congratulations, you've proven that there are at least 3 solutions for the equation.

Is n "tending" to infinity in the above ? I suppose so.

Why do you think I set up the limit? Because I don't simply "suppose so".

Do real solutions exist all the way regardless of the finite value of n, no matter how large? Obviously.

I never said they didn't.

Does it make sense to ask the question for an infinite value of n ? NO.

And here is where your logic breaks down.

 

[Icebreaker]

By your logic, suppose we have $$y=\frac{1}{x}$$. We find that there is a non-zero $$y$$ for any arbitrarily large $$x$$, and therefore makes no sense to ask whether there's a limit for $$y=\frac{1}{x}$$ as $$x\rightarrow\infty$$.

 

[Justin Lazear]

I never said n EQUALS infinity.

Does it make sense to ask the question for an infinite value of n ? NO.

And here is where your logic breaks down.

I'm confused... Do you want n to equal infinity or not?

--J

 

[Icebreaker]

No, I don't. Read the post above yours.

 

[DeadWolfe]

Could you state your meaning using a more rigorous definition of limits?

 

[robert Ihnot]

To Illustrate: a case where you could use the limit is the case of the Fibonacci numbers: 1,1,2,3,5,8,13,21,34,55,89,144...where the series in continued by the formula: F(N-1)+F(N)=F(N+1), or we add two successive numbers to get the next one.

Now it happens to be that this series is closely connected with the Golden Mean:
$$\frac{-1+\sqrt(5)}{2}$$ =.618034...

We look at: 34/55 =.618182..;55/89 =.617978...;89/144 =.618056...

Thus we might wonder if the limit as n goes to infinity of F(n)/F(n+1) = Golden Mean...which it does!

This limit is so good, that if divided 144 by the Golden Mean we get 232.996, which rounds off to 233 =89+144. (In fact it is good even from the second "1" giving 1/GM = 1.6180 rounds to 2. 2/GM =3.23 rounds to 3, 3/GM=4.85 rounds to 5!

 

[Justin Lazear]

Do real solutions exist all the way regardless of the finite value of n, no matter how large? Obviously.

I never said they didn't.

I'm trying to show the following:

Let $$a^n+b^n=c^n$$ where $$a, b, c$$ are not equal to zero. There is no real solution for $$a, b, c$$ as $$n\rightarrow\infty$$.

And I don't understand what the post that supposedly explains whether or not you want n to be infinity. I'm still confused on this issue.

--J

 

[Curious3141]

Oh? $$lim_{n\rightarrow\infty}\frac{1}{n}=0$$

Oh-what ? $$\frac{1}{n}$$ is an EXPRESSION, not an EQUATION. And it *never* equals zero no matter what value of n you choose. Only the limit is zero. There is a fundamental distinction here you don't seem to be grasping.

Of course there's a solution for every finite n. So?

So the premise you set about to "prove" is WRONG !

I'm investigating the value of the expression as a variable is increasing without bounds. And?

No, you are NOT. You are investigating an EQUATION as n tends to infinity. Which makes no sense.

You CAN investigate the behavior of the EXPRESSION $${(a^n + b^n)}^{\frac{1}{n}}$$ as n tends to infinity. If, for simplicity, we allow only for positive real values of a and b, then

$$\lim_{n \rightarrow \infty}{(a^n + b^n)}^{\frac{1}{n}} = max(a,b)$$

Fair enough ?

*If* you say $$c = {(a^n + b^n)}^{\frac{1}{n}}$$ THEN it's perfectly acceptable to say :

$$\lim_{n \rightarrow \infty}c = max(a,b)$$

Here you are still investigating the behavior of the expression called "c" as n tends to infinity. It's completely fine.

OTOH, it is not standard practice (and hence not acceptable) to do the same for an equation trying to find solutions at a limit.

 

[DeadWolfe]

I tihnk your problem is that you haven't defined "solution" well enough, when you said there wasn't one.

 

[Icebreaker]

Oh-what ? $$\frac{1}{n}$$ is an EXPRESSION, not an EQUATION.

Oh? And

$$\frac{a^n+b^n}{c^n}$$

is NOT an expression?

So the premise you set about to "prove" is WRONG !

Interesting how you can rush to that conclusion based on 3 examples where n equalled finite numbers.

No, you are NOT. You are investigating an EQUATION as n tends to infinity. Which makes no sense.[...]

No, I am investigating the EXPRESSION

$$\frac{a^n+b^n}{c^n}$$

as n increases without bound. And from what I see, there's little difference between what I am trying to do, and the example which you gave.

 

[Icebreaker]

I tihnk your problem is that you haven't defined "solution" well enough, when you said there wasn't one.

Perhaps. I will try and find a better definition of "solution".

And I don't understand what the post that supposedly explains whether or not you want n to be infinity. I'm still confused on this issue.

--J

No, you read the wrong post. Read the one directly above your first reply to this thread.

 

[Curious3141]

Let's take things one at a time...

Oh? And

$$\frac{a^n+b^n}{c^n}$$

is NOT an expression?

This IS an expression. If you have rigidly defined $$c^n = a^n + b^n$$ then the expression is ALWAYS equal to unity. The limit is also unity.

Going back to your first post :

$$lim_{n\rightarrow\infty} \frac{a^n+b^n}{c^n}=1$$

must also be true.[/tex]

Fine so far.

We can use properties of limits and algebra to obtain, from the previous equation

$$lim_{n\rightarrow\infty} \frac{a^n}{c^n}+lim_{n\rightarrow\infty} \frac{a^n}{c^n}=1$$

Also fine.

It can be shown that if $$a^n+b^n=c^n$$ (where $$a, b, c$$ are not equal to zero) is true, then $$c>a$$ and $$c>b$$

True for all finite n, but NOT at the limit. At the limit, $$c = max(a,b)$$. If b is the greater, then $$c = b$$ at the limit. This is your slip-up.

Therefore

$$0<\frac{a}{c}<1$$ and $$0<\frac{b}{c}<1$$

Again, true for all finite n cases, not correct at the limit.

Using properties of limits, we can state that

$$lim_{n\rightarrow\infty} \frac{a^n}{c^n} = lim_{n\rightarrow\infty} \frac{b^n}{c^n} = 0$$

Incorrect ! If b is the larger quantity, then $$lim_{n\rightarrow\infty} \frac{a^n}{c^n} = 0$$ while $$lim_{n\rightarrow\infty} \frac{b^n}{c^n} = 1$$

The rest is all wrong too.

and therefore

$$lim_{n\rightarrow\infty} \frac{a^n}{c^n}+lim_{n\rightarrow\infty} \frac{a^n}{c^n}=lim_{n\rightarrow\infty} \frac{a^n+b^n}{c^n}=0$$

Which is not 1, and we have thus shown that the contrary to the first statement cannot be true.

Interesting how you can rush to that conclusion based on 3 examples where n equalled finite numbers.

I'm sorry, but you do not understand limits.

 

[Curious3141]

You could have saved a lot of breath: my mistake was not in the way I am evaluating the limits, nor my understanding of limits, but circular logic. I assumed $$n\rightarrow\infty$$, but used a finite number to evaluate one of the properties I used in the proof, and replugged it back into my original guess.

Saved a lot of breath ? How about next time I don't bother to correct your misconceptions at all ?

And there IS something deeply wrong with your understanding of limits if you can make a mistake like that.

I'm done with this thread.

 

[Icebreaker]

Get it through your head: limits had nothing to do with it. It was circular logic.

 

[Curious3141]

Get it through your head: limits had nothing to do with it. It was circular logic.

GET IT THROUGH YOUR HEAD : Your entire attempt to find "solutions" to an "equation" at the limit was FLAWED. It wasn't a simple case of mistaking the behaviour of a finite property with the limit, IT WAS A COMPLETELY FLAWED ATTEMPT AT A PROOF BY CONTRADICTION !

FUNDAMENTALLY WRONG CONCEPT ! Not "circular logic". :yuck:

 

[Icebreaker]

My entire attempt to find "solutions" to the limits of both sides of the "equation" by contradiction was FLAWED because I mistook the behavior of a finite property with the limit.

Tell me what's wrong with the following, by your definition

$$x = x$$

$$lim_{x\rightarrow\infty}x = lim_{x\rightarrow\infty}x$$

 

[chronon]

Icebreaker's proof is perfectly valid. But all it shows is that there are only a finite number of values of n for which a^n+b^n=c^n for fixed a,b and c >0.

 

[matt grime]

I want to echo Chronon because Icebreaker was getting a lot of stick for someone else misunderstanding the question.

Rephrase it as consider the function f_n = x^n +y^n

and let f be the pointwise limit of f_n when this makes sense.
Are there any pairs x,y for which f(x,y)=1?

No, because the pointwise limit is identically zero for 0<x,y<1, 2 for x=y=1, and not defined other wise.

 

[Curious3141]

I want to echo Chronon because Icebreaker was getting a lot of stick for someone else misunderstanding the question.

Rephrase it as consider the function f_n = x^n +y^n

and let f be the pointwise limit of f_n when this makes sense.
Are there any pairs x,y for which f(x,y)=1?

No, because the pointwise limit is identically zero for 0<x,y<1, 2 for x=y=1, and not defined other wise.

Matt, look at the original question and the way it was phrased. Does it make sense ?

It was not the limit of an expression. It was an attempt to define real solutions to an *equation* with exponents as the exponents tended to infinity. As far as I'm aware, that is not sound mathematics.

Are you saying I'm wrong ? If so, please explain why.

 

[matt grime]

Yes, the original question makes sense as in it can certainly be understood even if it isn't the clearest way of writing it out: it defines a function as a pointwise limit. It would ghave been better to explain how it should have been better written since it is obvious what it is asking for.

How can you simultaneously say it isn't a limit and that things tend to infinity?

If you prefer, it is asking for the direct limit of the sets of solutions for each n ordered by inclusion - this makes it even clearer the answer is empty.

 

[Curious3141]

Yes, the original question makes sense as in it can certainly be understood even if it isn't the clearest way of writing it out: it defines a function as a pointwise limit. It would ghave been better to explain how it should have been better written since it is obvious what it is asking for.

No, it wasn't obvious to me what the question was asking for.

For me the solution to an equation has to satisfy the equation when it is substituted back into it. Agree ?

Now let me explain my problem with the question. If one tries to solve the equation algebraically "at the limit", the solution might just not work on the original equation.

Let's say we decide the values of a and b and want to determine what value of c (if any) exists. The obvious way of doing that is to evaluate $$c = {(a^n + b^n)}^{\frac{1}{n}}$$. It is clear that as n tends to infinity, the limit of $$c = max(a,b)$$. Fair enough ?

Now let's try putting that back into the original equation in the form $$a^n + b^n - c^n = 0$$, where n tends to infinity. Here we are evaluating each term sequentially. It is clear that no solution set other than the trivial (0,0,0) will actually work upon substitution.

I believe the original poster was trying to use this sort of logic to show that the equation can have no real solutions "at the limit". But to me, the question makes no sense. If you solve an equation with valid algebra, the solution *has* to work. Putting the solutions back in and verifying that they do not satisfy the equation does not mean that there are no real solutions to the equation per se, it means that the statement itself is flawed.

The equation (barring the limit) is in a valid algebraic form. The solutions that I found were with valid algebra. The problem was in the way that solution behaved when the limit was taken. That limit could not successfully be substituted back into the original equation.

The conclusion I draw from that is that it *makes no sense* to speak of solving an equation at the limit. It's fine to define the behaviour of a function at the limit, as you have done, but not try to solve an equation.

Do you still disagree, Matt ?

 

[Icebreaker]

Why does it "make no sense" to solve an equation at the limit?

$$x = x$$

$$lim_{x\rightarrow a}x = lim_{x\rightarrow a}x = a$$

Maybe you can explain your definition of "solving an equation" better.

 

[Curious3141]

Why does it "make no sense" to solve an equation at the limit?

$$x = x$$

$$lim_{x\rightarrow a}x = lim_{x\rightarrow a}x = a$$

Maybe you can explain your definition of "solving an equation" better.

Your notation is valid for finite a, but it breaks down for infinite a. In the case of a simple identity, the break-down is not obvious.

Try this :

$$lim_{x \rightarrow \infty}(x - 1) = lim_{x \rightarrow \infty}x$$

Does that imply that you've solved $$x -1 = x$$ ?

EDIT : Made it a much simpler example.

 

[Icebreaker]

No, it implies that as $$x\rightarrow\infty$$, the two expressions are equal. Without the infinite limit condition, the statement "the two expressions are equal" is false. Otherwise, it is true.

In other words, $$x - 1 = x$$ is true iff $$x\rightarrow\infty$$.

Or, if you prefer:

$$f(x)=f(x) \Rightarrow lim_{x\rightarrow\infty}f(x) = lim_{y\rightarrow\infty}f(y)$$

However

$$lim_{x\rightarrow\infty}f(x) = lim_{y\rightarrow\infty}f(y)$$ does not imply that $$f(x)=f(x)$$.

 

[Curious3141]

No, it implies that as $$x\rightarrow\infty$$, the two expressions are equal. Without the infinite limit condition, the statement "the two expressions are equal" is false. Otherwise, it is true.

I would prefer to say the expressions approach one another, and the limits of the expressions are equal. Now think : is that really an equation ? Have you truly "solved" for a real value of x ? Real numbers have to be finite.

By the same token, in your original problem statement, any real number set (a,b,c) that's put into the equation "at the limit" will be either reduced to zero or be increased without limit. From that you think you can say that "there is no real solution" to the equation at the limit (which is what I think you tried to prove).

But in truth, there are no real solutions to *any* (non-trivial) equations at infinite limits. This is why I feel that it makes no sense to even ask the question.

I just saw your edit :

In other words, $$x - 1 = x$$ is true iff $$x\rightarrow\infty$$

I disagree with this statement. I think you should phrase that as "the limit of x-1 is equal to the limit of x as x tends to infinity). Subtle difference, semantics perhaps, but to me, it makes a lot of difference.

 

[Icebreaker]

But in truth, there are no real solutions to *any* (non-trivial) equations at infinite limits. This is why I feel that it makes no sense to even ask the question.

What kind non-triviality are we talking about here? Because, as I pointed out before:

$$lim_{x\rightarrow\infty}\frac{1}{x}=a=0$$

And I disagree, it makes perfect sense to ask the question, even if you are right. Redundant, perhaps, but nothing illegal or illogical. After all, if you are right for any general case, then I can prove it for any special case, and I did. So what's the problem here?