Proving a=b=2: Simple yet Annoying Homework Equation Solution

[Norway]

Homework Statement

a+b = ab = a^b
Prove that a=b=2

Homework Equations

The Attempt at a Solution

Ok, I've tried this for more than an hour now, but can't figure out how to do it. I guess it could be done with Lambert's W-function (?) but this task is given on a much lower level. They haven't even learned how to prove that when x is a odd number, then x squared must be odd aswell. So this must be really simple, but still I can't figure anything out.

I've tried to take the equations and isolate a or b, and then insert it into one of the other functions, but I've gotten nowhere. I really hope someone could help me se the obvious here.

 

[icystrike]

This is part of the proof:
I'm in a hurry continue later(=

a+b\equiv0(moda)
b\equiv0(moda)
Therefore,a divides b

a^{b}\equivab(modb)
a^{b}\equiv0(modb)
Therefore,b divides a

Concluding that a=\pmb

 

[Hurkyl]

Don't give complete answers. But your approach can't work anyways, because it doesn't say anything about non-integers.


I presume that you're supposed to have an idea inspired by the qualitative behavior of addition vs. multiplication vs. exponentiation...

 

[ehild]

At low level I would mean to substitute the values given for a and b and see if the relations are true :)

ehild

 

[tiny-tim]

Hi Norway!

(try using the X² tag just above the Reply box )

Here's a start …

the bit on the left gives you an ab - a - b … does that remind you of anything?

 

[HallsofIvy]

At low level I would mean to substitute the values given for a and b and see if the relations are true :)

ehild

?? There were no "values given". The problem is to show that his is true for all numbers a and b.

 

[Norway]

the bit on the left gives you an ab - a - b … does that remind you of anything?

To be hones - no, it doesn't. :(
First thing that comes to mind are those (a+b)² rules, but I guess neither of them fits in here.
Then I think; "So a and b multiplied minus a and b is zero. That must mean that a+b=ab", and we're back to the start...
Sorry.

 

[tiny-tim]

Try adding a constant to it, and then factoring it.

 

[Norway]

Geez, I still don't get it.
I'm trying to look at this task like I have no idea what a and b could be, I'm not doing anything wrong there, am I?

Ok, ab-a-b=0. If we knew that a=b, then OK, but we don't. That's what we're going to prove. Right?

Or wait;
ab - a - b + 1 = 1 \ \Rightarrow \ (a-1)(b-1) = 1

Was this what you had in mind?
Then what? If they had to be integers, then sure, but they don't. Any more hints would be greatly appreciated. :D

 

[tiny-tim]

Or wait;
ab - a - b + 1 = 1 \ \Rightarrow \ (a-1)(b-1) = 1

Woohoo! ​

ok, now have a go at the bit on the right … ab = a^b …

how can you get an (a - 1) or a (b - 1) out of that?

 

[penguin007]

Hi everyone,
tiny-tim, I tried: ab=a^b
a-1=a^(b-1)-1 and since b-1=1/(a-1), then (a-1)(a^(1/(a-1))-1)=1...

but it's not a likeable equation,is it?
is there any other way?

 

[tiny-tim]

Hi penguin007!

(please use the X² tag just above the Reply box )

I managed to get as far as a^b-2b^a-2 = 1 …

not sure what to do then. ​

 

[icystrike]

Don't give complete answers. But your approach can't work anyways, because it doesn't say anything about non-integers.I presume that you're supposed to have an idea inspired by the qualitative behavior of addition vs. multiplication vs. exponentiation...

Hi Hurkyl!
You are true I've neglected the non-integers. But if it is non-integers, i do not know how should i deal with my proof.

Hi!Norway can you specify if a and b are real numbers or solely integers?

 

[Norway]

Hi! The task was given exactly as I gave it to you. I would guess this means all numbers, but maybe it was integers. As I said, this was given on a relatively low level, to students who has never ever done any proofs before.

Thank you guys.

 

[ehild]

Norway, was it said that "Prove that a=b=2" with the same words? Or was it meant like that: prove that a=b=2 satisfies the equations above? A simple substitution of a=2 and b=2 into the equations is a complete proof then, the derivation of the solution is not necessary. If the question means "prove that a=b=2 is the unique solution", that is a nice problem for high level, as this thread illustrates it, but not for beginners in maths.

ehild

 

[penguin007]

The expression you got seems to be interesting tiny-tim but I didn't manage to exploit it.

We can try to introduce the function f:=x->(x-1)*(x^(1/(x-1))-1) and then by studying the variations of this function, we would show that the equation (x-1)*(x^(1/(x-1))-1) =1 has a single solution (x=2, which would give us a=2); BUT:
1-I do not guarantee the study of f is easy;
2-this problem seems to be a more arithmetic one and therefore should be solved differently;

...

 

[Norway]

Norway, was it said that "Prove that a=b=2" with the same words? Or was it meant like that: prove that a=b=2 satisfies the equations above? A simple substitution of a=2 and b=2 into the equations is a complete proof then, the derivation of the solution is not necessary. If the question means "prove that a=b=2 is the unique solution", that is a nice problem for high level, as this thread illustrates it, but not for beginners in maths.

ehild

Well, it said "Given a+b=ab=a^b, prove: a=b=2"

But yes, I do start to wonder. They can express themselves quite sloppy sometimes, especially on lower levels. I just saw a good try on this on a Norwegian forum of maths (that's why I posted this here, I saw this question, and was unable to help, even though I'm on a much higher course ).

ab = a^b

(ab)^b = \left(a^b\right)^b

a^b b^b = a^{b \cdot b}

b^b = a^b

a = b

a+b = ab

2a = a \cdot a

a = b = 2

Obviously this turned out to be wrong, but I guess it was still a nice try.
They said they found a solution to this now, so I'm going to check out the link they gave and then check back here.

 

[Norway]

Okay, I just reviewed the other solution, and I can't see anything wrong with it. Maybe you guys can?

ab = a^b \ \Rightarrow \ b = a^{b-1}

a+b = ab \ \Rightarrow \ b = 1 + \frac{b}{a} \ \Rightarrow \ b - 1 = \frac{b}{a}

Therefore:

b = a^{b-1} = a^{\frac{b}{a}}

b = a^{\frac{b}{a}} \ \Rightarrow \ b^a = a^b

a \cdot \log{b} = b \cdot \log{a}

\frac{\log{a}}{a} = \frac{\log{b}}{b}

a = b
Is this correct here? Why not?

a+b = a^2 \ \Rightarrow \ 2a = a \cdot a \ \Rightarrow \ a = b = 2 \; \; \; Q.E.D.

 

[Norway]

\frac{\log{a}}{a} = \frac{\log{b}}{b}

a = b
Is this correct here? Why not?

It's not. Because you believed too much in the guy who said it was strictly increasing. If you actually take a look at the graph, you'll see that a=b could mean a ton (infinite?) of possible (a,b) pairs.

 

[tiny-tim]

Hi Norway!

That requires that logx/x be single-valued.

Unfortunately, it has a maximum value of 1/e at x = e (2.718…),

so loga/a = logb/b does not imply a = b.

EDIT: uhh? are you two different people (or countries)? ​

 

[Norway]

EDIT: uhh? are you two different people (or countries)? ​

Ha ha, no, I just replied to myself, because I asusmed the guy who posted this proof was telling he truth about it being strictly increasing, without even having looked at the function. Then I checked it out, and realized that it was wroong. Still a nice try, though.

 

[Norway]

Let me just ask you; is there anyone here who knows how to prove this at all?

 

[Hurkyl]

Here's the sketch of the first elementary complete proof I've put together.

I hope you'll allow me to assume a is positive, so that we don't have any worries about whether a^b is well-defined.


First solve ab=a^b (e.g. with logarithms): this gives

a = b^b-1​

Then a+b = ab gives

b^b-1 + b = b^b​

or equivalently

1 + b^2-b = b​

Checking the three cases b <= 1, 1 <= b <= 2, and 2 <= b yields b=2 is the only solution.

 

[tiny-tim]

Hi Hurkyl!

First solve ab=a^b (e.g. with logarithms): this gives

a = b^b-1​

No, a = b^1/(b-1).

 

[Hurkyl]

My first ideas simplest implementation, unfortunately, uses calculus to get the identity I wanted, and works out like this:

Solving a + b = ab gives

a = b/(b-1)​

Our assumption of a > 0 implies b > 1.

Plugging into ab = a^b gives

(b-1) log(b-1) = (b-2) log(b)​

If we set f(x) = (x-1) log(x-1) - (x-2) log(x), we see that the equation f'(x)=0 simplifies to

0 = 2/x + log (1 - 1/x)​

which has no solutions. Therefore f is monotone and has a unique zero on x>1: x=2.

NM I messed this one up too. I feel silly making two arithmetic errors in a row. Terrible ones even -- I shouldn't be allowed to do math today

 

[Hurkyl]

Hi Hurkyl!


No, a = b^1/(b-1).

Oh bah.

 

[vela]

Both a and b satisfy

x^{\frac{1}{x-1}} = \frac{x}{x-1}

if that helps. You can also show that 1&lt;a&lt;e.

 

[vela]

My first ideas simplest implementation, unfortunately, uses calculus to get the identity I wanted, and works out like this:

Solving a + b = ab gives

a = b/(b-1)​

Our assumption of a > 0 implies b > 1.

Plugging into ab = a^b gives

(b-1) log(b-1) = (b-2) log(b)​

If we set f'(x) = (x-1) log(x-1) - (x-2) log(x), we see that the equation f'(x)=0 simplifies to

0 = 2/x + log (1 - 1/x)​

which has no solutions. Therefore f is monotone and has a unique zero on x>1: x=2.

NM I messed this one up too. I feel silly making two arithmetic errors in a row. Terrible ones even -- I shouldn't be allowed to do math today

What's wrong with this one?

 

[Hurkyl]

What's wrong with this one?

0 = 2/x + log (1 - 1/x)​

has solutions. Or, at least, it doesn't obviously not have solutions. Somehow I convinced myself this was positive + positive = zero.

 

[Kaimyn]

I think I may have something. Correct me if I'm wrong:

ab=a^b which implies a=\sqrt[b-1]{b}
a+b=ab which implies a = b/(b-1)
Then b = b^b-1/(b-1)^b-1
From this we get: b^b-2 = (b-1)^b-1
Expanding gives: b^b-2 =b^b-1-_b-1C¹b^b-2+...
And _b-1C¹ = b-1
From this we get: 0 = b^b-1-(b-1+1)b^b-2+... = b^b-1-b^b-1

Now, correct me if I'm wrong, but I think this implies that there are only two entries in the (b-1)th row in pascals triangle, so b-1=1 which implies b = 2.

From there: ab = a^b implies 2a = a² implies a = 2

Is this valid?

 

[Hurkyl]

Now, correct me if I'm wrong, but I think this implies that there are only two entries in the (b-1)th row in pascals triangle, so b-1=1 which implies b = 2.

What is your rationale? I'm guessing you're implicit argument is that the rest of terms have to sum to zero, therefore they must individually be zero. But that argument clearly doesn't work -- it's easy to construct a counterexample. The argument would work if we had a sum of nonnegative terms, but that isn't the case.

Also, you've assumed b is an integer.

 

[Hurkyl]

Oh, bah, my calculus proof can be salvaged.

f''(x) < 0 on (2, +infty)

Therefore, f'(x) >0 on (2, +infty) (it decreases to zero)

Therefore, f(x) is increasing on (2, +infty) -- in particular it has no zeroes in that interval.

Therefore, b <= 2. Furthermore, this implies a >= 2.

By symmetry^*, we also deduce a <= 2 and b >= 2.

Therefore a=b=2.

*: Recall we've already shown a+b=ab=a^b=b^a

 

[Altabeh]

Oh, bah, my calculus proof can be salvaged.

f''(x) < 0 on (2, +infty)

Therefore, f'(x) >0 on (2, +infty) (it decreases to zero)

Therefore, f(x) is increasing on (2, +infty) -- in particular it has no zeroes in that interval.

Therefore, b <= 2. Furthermore, this implies a >= 2.

By symmetry^*, we also deduce a <= 2 and b >= 2.

Therefore a=b=2.

*: Recall we've already shown a+b=ab=a^b=b^a

You don't need to go further from (x-1)*log(x-1)-(x-2)*log(x)=f(x): This function is strictly increasing over [2,+oo) and the only solution to f(x)=0 is x=2=b (remember for x between 1 and 2 the function does not hit the x-axis). Thus plugging this into a+b=ab gives a=2. Period

AB

 

[Hurkyl]

You don't need to go further from (x-1)*log(x-1)-(x-2)*log(x)=f(x): This function is strictly increasing over [2,+oo) and the only solution to f(x)=0 is x=2=b (remember for x between 1 and 2 the function does not hit the x-axis).

Both of those facts must be argued, not just assumed.

I and found the symmetry argument simpler than arguing f(x) has no zeroes on (1,2).

 

[Altabeh]

Both of those facts must be argued, not just assumed.

I and found the symmetry argument simpler than arguing f(x) has no zeroes on (1,2).

You can check the first claim by a simple analytical comparision between terms:

1-(x-1)*log(x-1)-(x-2)*log(x)=f(x); Here we have two factors (x-1) and (x-2) s.t. for x>=2, the rate of change of (x-1) w.r.t. the change of x is clearly faster than that of (x-2) by a constant angle. Also the rate of growth of log(x-1) is so much slower than log(x) but one knows that all logarithm functions grow slower than linear functions, leading to a faster-growing term (x-1)*log(x-1) compared to (x-2)*log(x). So the function is strictly increasing on [2,+oo).

2- From 1, it is too easy to show that a strictly increasing function hits the x-axis once and only once. (Imagine how a line can only have ONE x-intercept.)

Anything difficult here so as to go for the second derivative of f, instead?

AB

 

[Hurkyl]

Anything difficult here so as to go for the second derivative of f, instead?

I don't know -- the argument is imprecise enough that I can't tell if you've said anything correct!

e.g. I'm not sure what you might mean by "the rate of change of (x-1) w.r.t. the change of x" such that it is both something different than the corresponding thing for (x-2), and the difference could be said to be a constant angle.


The advantage to the derivative is that it turns everything into simple arithmetic -- one can eliminate (or drastically reduce) the need for imprecision, and it's usually much easier to spot and check all of the edge cases.


e.g. I could have just asserted f'(x) = 2/x - log(1 - 1/x) is positive on (2,+infty). After all, log(1-1/x) looks like -1/x, which is half the size of 2/x! But there is the edge case to consider -- is 2 really close enough to +infty for us to rely on that approximation?

I could try to puzzle it out, but it's so much easier and clearer (both to myself and for the reader) to simply observe that f''(x) = -(x-2)/(x² (x-1)) is somewhat more obviously negative on (2, +infty).

 

[Altabeh]

I don't know -- the argument is imprecise enough that I can't tell if you've said anything correct!

e.g. I'm not sure what you might mean by "the rate of change of (x-1) w.r.t. the change of x" such that it is both something different than the corresponding thing for (x-2), and the difference could be said to be a constant angle.


The advantage to the derivative is that it turns everything into simple arithmetic -- one can eliminate (or drastically reduce) the need for imprecision, and it's usually much easier to spot and check all of the edge cases.


e.g. I could have just asserted f'(x) = 2/x - log(1 - 1/x) is positive on (2,+infty). After all, log(1-1/x) looks like -1/x, which is half the size of 2/x! But there is the edge case to consider -- is 2 really close enough to +infty for us to rely on that approximation?

I could try to puzzle it out, but it's so much easier and clearer (both to myself and for the reader) to simply observe that f''(x) = -(x-2)/(x² (x-1)) is somewhat more obviously negative on (2, +infty).

I think you got me wrong. If something sounds meaningless to you, I assume, it is not false or "imprecise" in any sense. What I meant is that, simply, while the slopes of x-1 and x-2 are the same, but the first one lies above the latter one, so x-1 always is greater than x-2 by a constant difference. (Maybe I was blurry or evern wrong on this one in my early post.) And about log(x-1) and log(x) which everything about the growth of f(x) depends only on 'em... these two for x>=2 are again in a similar situation as log(x) lies right above log(x-1) but not by a varying difference (as x-->oo log(x)=log(x-1)) and since they are "strictly increasing" as are x-1 and x-2, so by analogy, you can see that linear functions are faster than logarithms in their growth, so (x-1)*log(x-1) is ALWAYS larger than (x-2)*log(x) for x>=2, so strictly increasing. This, I think, would shed more light on the whole problem besides your mathematical reasoning.

Anyways, I must say that your idea of proof is brilliant...
AB

 

[Hurkyl]

Maybe that will click if I spend some time thinking it through.



I can't claim credit for discovering my method of proof -- I actually think it's a standard trick that, for some reason, isn't really made to sink in after your first calculus class.

It's useful enough that it's usually one of the first few things I try on a problem like this. Though, as it turns out, in this particular case, I computed everything because I was trying to "graph" the function. (I don't have a calculator handy)

I use scare quotes because I didn't actually plot it -- I just wanted broad strokes like "f''(x) is increasing from -infty to 0 over (2, +infty)".