- 21 Oct '07 15:33There are two positive numbers, A and B, which happen to have exactly the same decimals and in the same order as well.

Another property of these numbers is that when you invert one of them you get the other, and vice versa.

Is this information enough to decide which are these numbers? And if so, what are they? - 21 Oct '07 16:46Hm...

Okay so let's say [A] is the integer part of A and [B] the integer part of B. Then A= [A]+x and B=[B]+x where x is the fractional part of both as the condition states. Then form 1/([A]+x) = [B]+x we get [A][B]+([A]+[B])x+x^2 = 1

Since the numbers are positive [A].[B] is greater or equal to 1, unless either [A] or [B] is 0. Thus we have without loss of generality 1= [A]x+x^2 or

x^2+[A]x-1=0. The positive root of this equation is x=(-[A]+sqrt([A]^2+4))/2. Now we need to find for what integers [A] this becomes less than 1 (since x is a fraction). So we need to evaluate how does the function (-n+sqrt(4+n^2))/2 behave with the increasing of n. This would always less than 1 since sqrt(4+n^2) is less then n+2 thus the final result is less than 1 for all n besides 0.

Does that sound right? - 21 Oct '07 19:25 / 1 edit

Correct.*Originally posted by AttilaTheHorn***For some reason, these numbers don't print out in the forum. the numbers are**

the square root of 5 minus 1and divide that answer by 2. the reciprocal is 2 divided by (the square root of 5 minus 1).

these numbers are 0.618033989. . . and 1.618033989. . .

Let the numbers we are looking for be (A+x) & x where A is an integer and x is between 0 and 1.

Then we have

A+x = 1/x

x^2 + Ax = 1

(x + A/2)^2 = 1 + (A^2)/4

x = sqrt{1 + (A^2)/4} - A/2

x = 1/2[sqrt{4 + (A^2)} - A]

We can see that for A>1 x>1 therefore our only solution is A=1

substituting for A.

x=1/2[sqrt{5} -1]

Nice problem with unexpected solution. - 21 Oct '07 19:57 / 3 edits

How do we see that?*Originally posted by wolfgang59*

x = 1/2[sqrt{4 + (A^2)} - A]

We can see that for A>1 x>1 therefore our only solution is A=1

substituting for A.

For A=2 for example we get 1/2 (sqrt (8) -2)) = sqrt (8)/2 - 1.

Now sqrt (8) < 3 thus sqrt (8) /2 is less than 1,5.

Then x=sqrt(8)/2 - 1 which is less than 1,5-1 = 0,5 which is less than 1.

In fact for any integer larger than 0 we get an x smaller than 1.

EDITED: Grrr something wrong with "less" sign. - 21 Oct '07 20:32

Thanks to ilywrin the answer is NO. Insufficient info for a unique solution.*Originally posted by FabianFnas***There are two positive numbers, A and B, which happen to have exactly the same decimals and in the same order as well.**

Another property of these numbers is that when you invert one of them you get the other, and vice versa.

Is this information enough to decide which are these numbers? And if so, what are they? - 22 Oct '07 05:46wolfgang59 among others showed nicely the calculations leading to the numerical value by AttilaTheHorn among others. Well done!

Next question:

Calculate the value of the expression

sqrt(1+sqrt(1+sqrt(1+ ... )))

where ... stands for sqrt(1+ ... ) recursively

and sqrt stands for square root. - 22 Oct '07 06:52

I will try my best - not sure how "legal" is this but as a formal approach it might work. Let us denote x=sqrt(1+sqrt(1+(sqrt(1+...)))*Originally posted by FabianFnas***wolfgang59 among others showed nicely the calculations leading to the numerical value by AttilaTheHorn among others. Well done!**

Next question:

Calculate the value of the expression

sqrt(1+sqrt(1+sqrt(1+ ... )))

where ... stands for sqrt(1+ ... ) recursively

and sqrt stands for square root.

Now we can write sqrt(1+x)=x since the both continue to infinity (hopeful that this is okay). Thus we get again quadratic equation: x^2-x-1=0 or x=(1+sqrt5)x2^(-1) since the other root is obviously negative and x is strictly positive. Would that work as an answer? - 22 Oct '07 07:34

ilywrin seems to have got it but one of his logic jumps is not clear (at least not to me!) Also I am unclear as to why he discarded one solution of his quadratic?*Originally posted by ilywrin***I will try my best - not sure how "legal" is this but as a formal approach it might work. Let us denote x=sqrt(1+sqrt(1+(sqrt(1+...)))**

Now we can write sqrt(1+x)=x since the both continue to infinity (hopeful that this is okay). Thus we get again quadratic equation: x^2-x-1=0 or x=(1+sqrt5)x2^(-1) since the other root is obviously negative and x is strictly positive. Would that work as an answer?

Define x as ilywrin has done; then

x=sqrt[1+sqrt(..

square the equation

x^2 = 1+x

Solutions of which are x=(1+sqrt5)/2 and (1-sqrt5)/2 - 22 Oct '07 09:46

This is rubbish. I dont know why its rubbish but it is!*Originally posted by wolfgang59***ilywrin seems to have got it but one of his logic jumps is not clear (at least not to me!) Also I am unclear as to why he discarded one solution of his quadratic?**

Define x as ilywrin has done; then

x=sqrt[1+sqrt(..

square the equation

x^2 = 1+x

Solutions of which are x=(1+sqrt5)/2 and (1-sqrt5)/2

Firstly by playing around a bit I can come up with many 'solutions'. For example consider squaring the equation x^2 = x + 1 then making substitution for x .... Let y=x^2 and you get a quadratic in y with incompatible results.

BUT more importantly just LOKING at the original problem;

Take one term and x = sqrt(1+d) so x is greater than 1.

Take two terms and x = sqrt[1 + sqrt(1 + e)} so x is greater than 2.

Take 3 terms and ....

I can make it as big as I like!

I therefore believe that Fabian's original problem is invalid .. although I dont understand why! - 22 Oct '07 11:48Pretty much - what this problem is about is to find the limit of the sequence Kn, when n tends to infinity, and Kn = sqrt(1+(sqrt(1+...))) - n 1s total.

K(n+1)>Kn but with how much? Let us define t(n)=K(n+1) -Kn . If t(n) tends to 0 as n grows to infinity then Kn is a bounded sequence, and we know that it has a limit. Else Kn grows arbitrarily large with n.

But t(1)=sqrt (2) - 1; t(2)= sqrt(1+sqrt(2)) -sqrt(2); so we see that t(n) is a decreasing sequence. (one can prove this by induction). Since it has positive members it would tend to 0 when n tends to infinity.

Then Kn is bounded and increasing.

Thus I believe we can use the formal method. As to wolfgang's remark - we discard the negative root since the sequence has positive members, hence must have a positive limit. - 22 Oct '07 15:16

This is the rubbish post. When you needlessly complicate the problem by squaring both sides of x^2 = x + 1 and substitute in y = x^2, you don't get a quadratic because it contains a square root sign. Secondly, your logic with the successive square roots doesn't follow. It's true that if "d" is a positive real number, then SQRT(1+d) >= 1, but not true that if "e" is a positive real number then SQRT[1 + SQRT(1+e)] >= 2. For example, let "e" be 0.21, then SQRT(1+e) = 1.1, and SQRT(1+1.1) =~ 1.449 which is less than 2.*Originally posted by wolfgang59***This is rubbish. I dont know why its rubbish but it is!**

Firstly by playing around a bit I can come up with many 'solutions'. For example consider squaring the equation x^2 = x + 1 then making substitution for x .... Let y=x^2 and you get a quadratic in y with incompatible results.

BUT more importantly just LOKING at the original problem;

Take one ...[text shortened]... herefore believe that Fabian's original problem is invalid .. although I dont understand why!

The original problem is quite valid, and ilywrin has already posted the correct solution. - 22 Oct '07 17:45

OK! My post rubbishing my post is also rubbish.*Originally posted by PBE6***This is the rubbish post. When you needlessly complicate the problem by squaring both sides of x^2 = x + 1 and substitute in y = x^2, you don't get a quadratic because it contains a square root sign. Secondly, your logic with the successive square roots doesn't follow. It's true that if "d" is a positive real number, then SQRT(1+d) >= 1, but not true that if ...[text shortened]... e original problem is quite valid, and ilywrin has already posted the correct solution.**

[hands held aloft & waving white flag!]