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 $1<a<e$.

 

[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?