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Old 2012-11-23, 21:43   #1
f1pokerspeed
 
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Question (xy)^2 when x and y are powers of two is solved!!

I have found some weird way of expressing the expression (redundant, he he) (xy)2, and I have no idea as to why it works when x and y are powers of two - see the image for the conjecture and expressions:
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Last fiddled with by f1pokerspeed on 2012-11-23 at 21:46
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Old 2012-11-23, 22:35   #2
NBtarheel_33
 
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Please recheck your notation. In particular, if x and y are both perfect powers of 2, they will be integers. You seem to be saying that (xy)^2 = d /xy^3, which would mean that the LHS (an integer) is equal to the RHS (a rational quantity, not always an integer). Your notation seems confused. Think about what you are trying to say, and recast it more clearly. Also consider trying some small examples for x and y.
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Old 2012-11-24, 00:38   #3
science_man_88
 
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Quote:
Originally Posted by NBtarheel_33 View Post
Please recheck your notation. In particular, if x and y are both perfect powers of 2, they will be integers. You seem to be saying that (xy)^2 = d /xy^3, which would mean that the LHS (an integer) is equal to the RHS (a rational quantity, not always an integer). Your notation seems confused. Think about what you are trying to say, and recast it more clearly. Also consider trying some small examples for x and y.
d=sqrt(2^d) leads to d=2 or d=4 checking integers under a million, regardless of x or y. (xy)^2 = 2^2*(log(x)+log(y)) when x and y are powers of 2. my math puts 2^x-2^y = 2*((2^(x-1)-1)-(2^(y-1)-1)).

Last fiddled with by science_man_88 on 2012-11-24 at 00:39
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Old 2012-11-24, 13:50   #4
NBtarheel_33
 
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Stepping through this a little further...

"When x and y are a perfect power of two" implies that x = 2^k and y = 2^m for some nonnegative integers k, m. Without loss of generality, let us make k >= m, i.e. make x >= y.

"The following formula can be used as a generalization to solve the expression (xy)^2". Not sure what "generalization" means in this context; perhaps the simple word "method" might be a better fit. And expressions cannot be "solved". But *equations* can be solved. Expressions are generally *evaluated*.

In any event, you claim that (xy)^2 = 1/d * (y^2) + xy, where "d is the difference in the powers of x and y, i.e. 2^d = 2^x - 2^y and d = SQRT(2^d)." This is a notational nightmare, and essentially obliterates any sort of meaning from the problem as you have stated it. Let's try to salvage some sensibility. I assume that for x = 2^k and y = 2^m, with k >= m, you want d = k - m = log2(x) - log2(y).

Now y^2 = (2^m)^2 = 2^(2m) by the rules for exponentiating exponents. Moreover, xy = (2^k)(2^m) = 2^(k+m) by the rules for multiplying exponents. Hence your claim that

(xy)^2 = 1/d * (y^2) + xy

boils down to

(2^(k+m))^2 = 1/d * 2^(2m) + 2^(k+m), or

2^(2k+2m) = 1/d * 2^(2m) + 2^(k+m).

Subtracting 2^(k+m) from both sides would yield

2^(2k+2m) - 2^(k+m) = 1/d * 2^(2m).

Next, factor the left hand side to get

[2^(k+m)] * [2^(k+m) - 1] = 1/d * 2^(2m).

Remember that k >= m. Therefore, k + m >= 2m, so that we may divide both sides of the above equation by 2^(2m), yielding by properties of division of exponents

[2^(k+m-2m)] * [2^(k+m) - 1] = 1/d, or

[2^(k-m)] * [2^(k+m) - 1] = 1/d.

Now recall that k, m are nonnegative integers. That means that the left hand side of the above equation is a nonnegative integer. More importantly, it also means that the *right hand side*, namely 1/d must also be a nonnegative integer. The only time this occurs is for d = 1, which implies 1/d = 1. But then we must have both of the factors on the left hand side equal to 1, which is impossible because the factors are of different parity (one is a power of two, hence even; the other is one less than a power of two, hence odd). We have thus reached a contradiction, and the stated "identity" must therefore be incorrect.

*Q. E. D.*

As a side note, writing statements such as d = SQRT(2^d), where a variable name is used not only for more than one quantity, but recursively in its own definition(!), is a huge faux pas in mathematics. Definitely something to avoid. Think about this: Suppose d = 5. Then you are saying that 5 = SQRT(2^5) = SQRT(32)!
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Old 2012-11-24, 21:52   #5
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The only possibilities for "d" are 2 and 4 because sqrt(22)=sqrt(4)=2 and sqrt(24)=sqrt(16)=4.

As the OP said, (xy)2=y2/d+xy. If x=2m and y=2n then it becomes 22m+2n=22n/d+2m+n and simplifying results in 4m+n=4n/d+2m+n

Last fiddled with by Stargate38 on 2012-11-24 at 22:38
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Old 2012-11-24, 22:47   #6
f1pokerspeed
 
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Thanks for all of your input on this, and I think I get the hang of what is being said here. Sorry about all of the flaws that there are in my original post, I'm kinda new to writing in mathematical notation (in this context, anyways.)
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Old 2012-11-25, 00:05   #7
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Dividing both sides by 2n results in 4m2n=2n/d+2m. Substituting 2 or 4 for d gives the following 2 possibilities:

4m2n=2n-1+2m (d=2)
4m2n=2n-2+2m (d=4)

Dividing both sides by 2n again gives:

4m=1/2+2m-n (d=2)
4m=1/4+2m-n (d=4)

since 4=22, the equations become:

22m=1/2+2m-n (d=2)
22m=1/4+2m-n (d=4)

Last fiddled with by Stargate38 on 2012-11-25 at 00:09 Reason: insert "d" values
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