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June 17th, 2009, 10:58 AM   #1
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LCD type problem

I have 2 real numbers m and n. I need to find a number p such that m * p and n * p both result in a whole number.

Seems like this should be pretty easy but the formula for this is eluding me at the moment.

Thanks.
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June 17th, 2009, 11:22 AM   #2
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The easy answer is to use p = 0. If p has to be non-zero, you can still succeed if (and only if) m and n are rational, i.e., if m = a/b and n = c/d, where a, b, c, and d are integers. You can then use p = b * d (and there will be other values for p that will also work).
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June 17th, 2009, 12:14 PM   #3
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Re: LCD type problem

p must be non-zero.

m and n are rational numbers.

I have no guarantee though that given m = a/b that a and b are integers (same with n = c/d). e.g, in my specific sample case I have m = 0.128 (or 1 / 7.8125) and n = 0.05 (1 / 20)
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June 17th, 2009, 12:32 PM   #4
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Re: LCD type problem

If it's any help 0.128 = 16/125
500 * 16/125 = 64, 500 * 1/20 = 25.
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June 17th, 2009, 01:05 PM   #5
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Re: LCD type problem

Let p be the LCM of the denominators of m and n in their simplest form.
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June 17th, 2009, 01:45 PM   #6
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If a non-zero p exists such that mp = q, an integer, it follows that m = q/p, so m can be expressed as a quotient of two integers. If m is originally given in some other form, such as 0.128, just rewrite that as 128/1000 or 16/125 or any other similar quotient having the same value. If, however, m = ?2 or ln(2) (or any other irrational value), it is impossible to express it as a quotient in that way, and p can't be found.
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June 17th, 2009, 02:14 PM   #7
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Re:

Quote:
Originally Posted by skipjack
If a non-zero p exists such that mp = q, an integer, it follows that m = q/p, so m can be expressed as a quotient of two integers. If m is originally given in some other form, such as 0.128, just rewrite that as 128/1000 or 16/125 or any other similar quotient having the same value. If, however, m = ?2 or ln(2) (or any other irrational value), it is impossible to express it as a quotient in that way, and p can't be found.
This is sufficient to solve my problem as I can put a limit on the number of significant digits. Thanks.
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