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Q&A #3962 |
From: Teh Oh Kian
To: Teacher2Teacher Public Discussion
Date: 2000122513:01:45
Subject: Re: Re: Re: A special linear equation
Thanks for helping to solve. I believe your solution is perfectly correct to prove that the set of feasible solutions is countably infinite, but leaving a doubt that in your solution, both the x and y are even or odd. According to definition of divisibility, integers can either be positive or negative. According to the comncepts of multiples, we say that a is a multiple of b if there exists a positive integer c so that a = b* c. Since any integer is a multiple of itself, my solution covers the case that either x or y is even but not both. From both our solutions, there exist at least two distinct sets of feasible solutions and both are countably infinite. Take the first set as {(k, k+1)} and the second as {(k, k+2)}[Note, in your solution, we can use k and k+2 instead of k-1 and k+1, afterall the difference must be 2.] Now my problem is to prove or disprove the existence of infinite number of such solutions. Now, if we were to investigate the set of cartesan products {(k, k+3)}, we can easily check that it is a set of feasible solutions whenever k is a multiple of 3. So, I conjecture that there exist a countably infinite increasing sequence of solutions {(k, k+r)}to such problems proposed earlier. Tahnks for the response, and I sincerely hope that more people, regardless of the ethnic groups, will contribute their brilliant ideas to enlighten me in my meagre know-how in Mathematics. Oh Kian
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