kmerfeld Posted March 11, 2013 Posted March 11, 2013 (edited) Hello everyone, I want to prove that for postive integers x, y, y is not equal to x+y. I want to do this using the WOP. Here's what I have done so far: Suppose for some postive integer x, there exists a y st y y+x. By the WOP, there exists a smallest x0 st y=x0+y. Now I think I may have to apply the WOP again, but am not sure. Any advice? Thanks a lot, Kevin P.S. A little help using special BBC code. I'd like to use LaTex but when I try to do the not equal sign it shows up as: [latex]\ne[/latex] How do I get rid of the < br > ? Edited March 11, 2013 by kmerfeld
Klaynos Posted March 11, 2013 Posted March 11, 2013 [latex]\ne[/latex] How do I get rid of the < br > ? Remove the return before the closing tag. 1
kmerfeld Posted March 11, 2013 Author Posted March 11, 2013 (edited) [latex]\ne[/latex] There she goes. Thanks Edited March 11, 2013 by kmerfeld
Nehushtan Posted March 13, 2013 Posted March 13, 2013 Hello everyone, I want to prove that for postive integers x, y, y is not equal to x+y. I want to do this using the WOP. Here's what I have done so far: Suppose for some postive integer x, there exists a y st y y+x. By the WOP, there exists a smallest x0 st y=x0+y. Now I think I may have to apply the WOP again, but am not sure. Any advice? Apply the WOP to [latex]y[/latex], not [latex]x[/latex]. First, however, show that there does not eixst any positive integer [latex]x[/latex] such that [latex]1=1+x[/latex]. For this you can make use of the axiom that 1 is not the successor element of any positive integer.
kmerfeld Posted March 13, 2013 Author Posted March 13, 2013 Hello Nehushtan, Thanks for the suggestion. However, the book requires that I first apply the WOP on x. Then it "hints" that I should apply the WOP again... KM
kmerfeld Posted March 14, 2013 Author Posted March 14, 2013 I think I figured it out: I want to show [latex]y \ne x+y [/latex] for all positive integers x and y. So suppose there exists a y such that [latex]y=x+y[/latex] Then by WOP, there exists a smallest x such that: [latex]y=x_{0} +y[/latex] Again by WOP, there exists a smallest y such that: [latex]y_{0}=x_{0}+y_{0}[/latex] Then: [latex]y_{0}-y_{0}=x_{0}[/latex] [latex]0=x_{0}[/latex] But 0 is not a positive number. So this is a contradiction, meaning: [latex]y \ne x+y [/latex] for all positive integers x and y. Now it does seem like I could have done this proof without using WOP, but it still seems to work this way, and this is how I was asked to do it. Thanks everyone. Consider question this solved.
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