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You are not logged in. #1 20070116 12:54:55
prove when a is a real number if a > 0 then 1/a > 0.I'm having a hard time proving this. #2 20070116 13:11:16
Re: prove when a is a real number if a > 0 then 1/a > 0.It all depends how much you're allowed to assume. Most of the time, such a proof is done without having to prove the basic results of inequalities, for example, if a > 0, a^2 > 0. Or even more basic, if a > 0, and b > 0, ab > 0. But this may not be the case. Anyways, here is a proof assuming you don't have to show the basic properties of inequalities. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #3 20070116 13:18:33
Re: prove when a is a real number if a > 0 then 1/a > 0.Consider this... #4 20070116 13:23:09
Re: prove when a is a real number if a > 0 then 1/a > 0.
Are you really allowed to say that here though? It seems to me that all this does is assume 1/a² > 0, which is the same concept as assuming 1/a > 0. It looks like you are using the statement we need to prove to prove the statement, if you know what i mean... kind of like defining a word and using the word in your definition. #5 20070116 13:26:06
Re: prove when a is a real number if a > 0 then 1/a > 0....however I do understand now; for any x ∈ R, x² > 0. You simply have x = 1/a. Your proof is fine. #6 20070116 13:47:37
Re: prove when a is a real number if a > 0 then 1/a > 0.Well, not exactly. You would just first have to prove that: "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #7 20070123 00:59:02
Re: prove when a is a real number if a > 0 then 1/a > 0.
But then you could just say that 1 > 0 and a > 0, so 1/a > 0. I think a problem like the one posed is not likely to let you assume that, since that would make the problem almost trivial. Bad speling makes me [sic] #8 20070123 03:05:42
Re: prove when a is a real number if a > 0 then 1/a > 0.*shrug* "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #9 20070123 07:22:13
Re: prove when a is a real number if a > 0 then 1/a > 0.The 'solution' / hint in the back of the book says Last edited by efmchico (20070123 07:46:56) #10 20070327 11:01:11
Re: prove when a is a real number if a > 0 then 1/a > 0.I just saw this thread for the first time while browsing around. So let me add my comments. #11 20070327 11:11:24
Re: prove when a is a real number if a > 0 then 1/a > 0.Jane, that seems to be a field theory approach to the proof. Nothing wrong with it of course, but when I read the question, I thought of it more as an analysis question. When dealing with analysis, you tend to just assume the general properties of all fields, even if they are theorems and not given by the definition of a field. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #12 20070327 11:25:04
Re: prove when a is a real number if a > 0 then 1/a > 0.Well, as you said, it depends on exactly what you can assume – so I’m making clear exactly what is assumed when dealing with real numbers; in this case, it’s the axioms for an ordered field. #13 20070327 11:30:54
Re: prove when a is a real number if a > 0 then 1/a > 0.
I entirely agree.
As much as possible, please. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." 