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June 17th, 2009, 10:58 AM  #1 
Newbie Joined: Jun 2009 Posts: 3 Thanks: 0  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. 
June 17th, 2009, 11:22 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,963 Thanks: 1606 
The easy answer is to use p = 0. If p has to be nonzero, 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).

June 17th, 2009, 12:14 PM  #3 
Newbie Joined: Jun 2009 Posts: 3 Thanks: 0  Re: LCD type problem
p must be nonzero. 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) 
June 17th, 2009, 12:32 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,807 Thanks: 1045 Math Focus: Elementary mathematics and beyond  Re: LCD type problem
If it's any help 0.128 = 16/125 500 * 16/125 = 64, 500 * 1/20 = 25. 
June 17th, 2009, 01:05 PM  #5 
Senior Member Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0  Re: LCD type problem
Let p be the LCM of the denominators of m and n in their simplest form.

June 17th, 2009, 01:45 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 18,963 Thanks: 1606 
If a nonzero 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.

June 17th, 2009, 02:14 PM  #7  
Newbie Joined: Jun 2009 Posts: 3 Thanks: 0  Re: Quote:
 

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