Mathematics desk
< May 9	<< Apr \| May \| Jun >>	May 11 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

May 10

Tried proving it - didn't work

This is slightly arbitrary.

I was given an A4 envelope from an accountancy firm. On the front of the envelope was the following expression.

$n(1+x)^{5}=2n$

Is there any way of proving that this expression is true or false? I tried, but didn't get anywhere. Dividing both sides by n and then 5th rooting each side, I was able to work out the value of x.

$n(1+x)^{5}=2n$
$(1+x)^{5}=2$
$1+x=1.148698355$
$x=0.148698355$

However, when using the above value of x to work out a value for n, the answer I end up with it 2n = 2n

Ai y da 20:24, 10 May 2007 (UTC)[reply]

I'm not sure what you're trying to prove - the statement isn't always true or always false; it's true or false depending on what the value of x and n are. Thus, you can solve for x and/or n to find what values make it true. What you've done is to find a value of x for which the statement is true for all n. (When x=0.148698355, the statement is always true, regardless of what the value of n is, since 2n=2n is true for all values of n.) There's also a value of n for which the statement is always true, regardless of the value of x - I'll leave you to find that on your own. Chuck 20:32, 10 May 2007 (UTC)[reply]

n(1+x)^5 = 2n

(1 + x)^5 = 2 provided n ≠ 0

5 possible solutions

x = 0.1487 provided n ≠ 0

x = -1.9293 - 0.6751i provided n ≠ 0

x = -1.9293 + 0.6751i provided n ≠ 0

x = -0.6450 - 1.0925i provided n ≠ 0

x = -0.6450 + 1.0925i provided n ≠ 0

Plus of course

x = anything and n = 0

202.168.50.40 03:19, 11 May 2007 (UTC)[reply]

Exponentiation with compass and straightedge

Can this be done? If so, how? Thanks --87.194.21.177 23:00, 10 May 2007 (UTC)[reply]

No. The only operations possible with Compass and straightedge are addition, subtraction, multiplication, division and square roots. -- Meni Rosenfeld (talk) 23:27, 10 May 2007 (UTC)[reply]

I thought there were only four operations, though I knew Descartes wasn't sure if they were four or five (because of the square root). His exact quote (well, translated from French): "Just as arithmetic consists of only four or five operations, namely, addition, subtraction, multiplication, division, and the extraction of roots, which may be considered a kind of division..." A.Z. 02:37, 11 May 2007 (UTC)[reply]

This edit is a test. A.Z. 22:38, 23 September 2007 (UTC)[reply]

So, considering the operations at hand, if an exponent has a finite base two representation, then, sure, it can be done. Root⁴(one) 02:52, 11 May 2007 (UTC)[reply]

In case I need to explain... Say you need to exponentiate

m

to binary 1101.0011 (13.1875 to us in decimal land), you'd find:

m^{8},m^{4},m,{\sqrt {\sqrt {\sqrt {m}}}},{\sqrt {\sqrt {\sqrt {\sqrt {m}}}}}

, multiply them all together (which adds the exponents), and you have your

\,m^{13.1875}

.

Root⁴(one) 03:15, 11 May 2007 (UTC)[reply]

This is clever, but it steps outside the system. When we say we can add two numbers, the two numbers are given inside the system as lengths, and we produce a length as the result. If we are asked to compute β^α, where both numbers are given as lengths, we cannot do so. --KSmrq^T 05:08, 11 May 2007 (UTC)[reply]

What if we were given three lengths: α, β and 1. Could we not then use 1 (and various divisions and multiplications of it) to measure the other two? (Don't you have to have a 1-length to construct a square root anyway?) - Rainwarrior 06:32, 11 May 2007 (UTC)[reply]

Yes, we are assuming that a 1 length is given. However, if I understand correctly what you meant by "use 1... to measure the other two", that is only possible if they are rational. If you allow any construction starting from 1, you can reach α and β iff they are constructible. -- Meni Rosenfeld (talk) 08:33, 11 May 2007 (UTC)[reply]

If you have a and b as lengths, you can construct the length (ab)^1/2, so if a unit length is given, you can do square roots in general. However, in the standard interpretation of compass-and-straightedge construction, no unit length is given, and then it is not possible to construct one. From the simple mantra "addition, subtraction, multiplication, division and square roots" it would appear possible (a/a), but actually multiplication and division are together one ternary operation: from a, b and c, construct a×b/c. --Lambiam ^Talk 11:05, 11 May 2007 (UTC)[reply]

(kill indent) So, just be sure, if I have a unit length "1" and a length "m" based off that unit length, I can essentially construct m^p for the restrictions of p I gave above, Correct? However it is not possible for not for me only possessing arbitrary lengths "m" and "n". I suppose I didn't factually answer the original question.

We've had a discussion on compass constructions within the last few months. I swear Compass and straightedge constructions#Constructible points and lengths is just as confusing as it was then. Is there not some way like:

If your number of interest β is constructible by recursively applying any of the following formula on measurements/lengths $\{m,n,p\}$ ~~previously constructed~~.

$m'$ <- $m+n$
$m'$ <- $m-n$
$m'$ <- $(mn)^{1/2}$
$m'$ <- $mn/p$
... other formulas...

Then you can also construct a length measuring β by compass and straight edge constructions. Or would that task be arduous, a long winded and almost useless list, not necessary, or original research? Thoughts? Root⁴(one) 15:57, 11 May 2007 (UTC)[reply]

Obviously, such a list would have to use better wording than above. Root⁴(one) 16:00, 11 May 2007 (UTC)[reply]

I think you are right about exponentiation. Given u and v, you can construct u^1−pv^p for those values of p that have a terminating binary expansion. Setting u := 1 gives the desired result. In general, if C(x₁, ..., x_n) is an expression in n variables using only the operations of addition, subtraction, multiplication, division and square root, then, given one length u and n lengths v₁, ..., v_n, the length uC(v₁/u, ..., v_n/u) is constructible. (Warning: I have not actually proved this.) The preceding statement for exponentiation is an application of this for C(x) = x^p. A clarification as suggested may be in order. Is there no authoritative source that has a concise and clear formulation? But further discussion of this should be taken to Wikipedia talk:WikiProject Mathematics and/or Talk:Compass and straightedge constructions. --Lambiam ^Talk 17:37, 11 May 2007 (UTC)[reply]