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Q&A #3962 |
From: Teh Oh Kian
To: Teacher2Teacher Public Discussion
Date: 2002052307:02:26
Subject: Re: Re: Re: Re: A special linear equation
It's much regret that I had not received any reply after 2 years to my conjecture about the existence of countably infinite increasing sequence {(k, k+r)} to a simple maths problem. I sincerely hope that there are beautiful people around to help me to solve this conjecture. I post the prvious discussions below. ...................................................................... From: Teh Oh Kian <mr_teh@hotmail.com> To: Teacher2Teacher Public Discussion Date: 2000121012:28:36 Subject: A special linear equation I was told by my State Eduction Director of Education that a teacher failed to write down the mathematical expression of the following word problem: Form an equation of two numbers whose sum is the multiple of their difference. The problem is actual simple. I tested on my son and he gave me the equation; x + y = k (x - y). It's perfectly correct. On further asking him to solve this equation, he hesitated. It is trivial that x and y are positive integers with x > y. On further investigation, the lower bound of y is 1 and the lower bound of x is 2. My problem is to prove or disprove that the set of solutions (x, y) is countably finite. If the set of feasible solutions is countably finite, find the upper bound of x and the upper bound y. From: Teh Oh Kian <mr_teh@hotmail.com> To: Teacher2Teacher Public Discussion Date: 2000121313:22:17 Subject: Re: A special linear equation In reviewing my own quetsion, can I let x = y + 1, y is an integer equal or greater than 1 as a particular solution? Since the set of positive integers is infinite, therefore there exist countably infinite feasible solutions of (x, y). Can anyone please help me to check whether this is a correct method. thanks. From: Pat Ballew <poetsoutback@yahoo.com> To: Teacher2Teacher Public Discussion Date: 2000122122:41:35 Subject: Re: Re: A special linear equation I will proceed with your assumption that x>y>0 although I see no reason to exclude the negatives in this problem (probably an oversight on my part).... but Given that x+y = k(x-y) we proceed to x+y = kx-ky which leads to ky+y = kx-x and (k+1)y= (k-1)x k-1 to produce y= ------ x (k+1) at this point we see that for every value of k, we can let x = k+1, and y= k-1 and we get a solution. Since this will work for all k>1 you have your infinite solution set. Hope I wrote that correctly... Pat Ballew, Misawa, Jp From: Teh Oh Kian <mr_teh@hotmail.com> To: Teacher2Teacher Public Discussion Date: 2000122514:01:45 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 concepts 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. Thanks 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 ...................................................................... I do hope the Webmatsers will at least help me to get people to solve the problem. Thanks. Oh Kian.
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