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**efmchico****Member**- Registered: 2007-01-15
- Posts: 7

I'm having a hard time proving this.

prove when a is a real number if a > 0 then 1/a > 0.

It seems so obvious, but i don't see a way to write a formal proof. Maybe by contradiction, but I still can't see how to write it.

Any help is appreciated

Thanks

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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.

Let a > 0. Then a^2 > 0. Thus, a/a^2 > 0. So a/a^2 = 1/a > 0.

"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..."

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**Zhylliolom****Real Member**- Registered: 2005-09-05
- Posts: 412

Consider this...

Since 1/a is the multiplicative inverse of a, it holds that a × 1/a = 1. Since a > 0, we have a positive number times the "unknown" status of 1/a equalling the positive number one. But a positive number must be multiplied by another positive number for the product to also be positive, so it must be that 1/a > 0.

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**Zhylliolom****Real Member**- Registered: 2005-09-05
- Posts: 412

Ricky wrote:

...Thus, a/a^2 > 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.

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**Zhylliolom****Real Member**- Registered: 2005-09-05
- Posts: 412

...however I do understand now; for any x ∈ R, x² > 0. You simply have x = 1/a. Your proof is fine.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Well, not exactly. You would just first have to prove that:

If a > 0 and b > 0, then a/b > 0.

Which is what I would consider a basic property 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..."

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**Dross****Member**- Registered: 2006-08-24
- Posts: 325

Ricky wrote:

Well, not exactly. You would just first have to prove that:

If a > 0 and b > 0, then a/b > 0.

Which is what I would consider a basic property of inequalities.

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]

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

*shrug*

A lot of times, inequality proofs such as these are fairly trivial.

"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..."

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**efmchico****Member**- Registered: 2007-01-15
- Posts: 7

The 'solution' / hint in the back of the book says

if a > 0 and 1/a <= 0

then 1= a*(1/a) <= a*0 =0.

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Can anyone decipher what is going on here? Is this a contradiction proof? Is it assuming a>0 and 1/a<=0 (which would be not 1/a >0)

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So does this sound like a contradiction proof,

assume a>0 and 1/a <= 0.

then 1/a = [a(1/a)] / a <= 0.

so a*(1/a) <= 0. (multiplying both sides by a)

but this contradicts our assumption that a > 0. (since a* 1/a is assumed greater than 0.)

Therefore, 1/a > 0.

----------------------------------

Thanks for feedback

*Last edited by efmchico (2007-01-22 08:46:56)*

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

I just saw this thread for the first time while browsing around. So let me add my comments.

The way to go about proving this is through the axioms for the real numbers. True, the real numbers can be built up from the rational numbers, which can be built up from the integers, which in turn can be built up from the natural numbers by the Peano axioms. (And you can go back even further by building the Peano axioms from set-theoretic axioms, particularly the axiom of infinity.) However, in an elementary textbook on analysis, one could bypass the complications and just start by axiomatizing the real numbers as a complete ordered field. (In other words, one merely takes the set of real numbers as given, defined by set of axioms which together uniquely characterize it.) The way to prove that *a* > 0 ⇒ *a*[sup]−1[/sup] > 0 is then through the axioms for an ordered field.

First, note that (−*x*)*y* = *x*(−*y*) = −(*xy*). These can be easily proved using the distributive axiom. We then use the following axioms for an ordered field: (i) If 0 ≤ *x* and 0 ≤ *y*, then 0 ≤ *xy*. (ii) If *x* ≤ *y*, then *x*+*z* ≤ *y*+*z*.

Assume to the contrary that 0 < *a* but *a*[sup]−1[/sup] ≤ 0, i.e. 0 ≤ −*a*[sup]−1[/sup]). Then 0 ≤ *a*·(−*a*[sup]−1[/sup]) = −1. But 0 ≤ −1 and 0 < *a* ⇒ 0 ≤ (−1)*a* = −*a* ⇒ *a* ≤ 0. This contradicts the fact that 0 < *a*. Hence it cant be true that *a*[sup]−1[/sup] ≤ 0; ∴ *a*[sup]−1[/sup] > 0. (Note: An ordered field is a totally ordered set and so the law of trichotomy for totally ordered sets applies: Given any *x*. *y*, exactly one of the following holds: *x* < *y*, *x* = *y*, *x* > *y*.)

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Well, as you said, it depends on exactly what you can assume so Im making clear exactly what is assumed when dealing with real numbers; in this case, its the axioms for an ordered field.

Anyway, as this is in the Euler Avenue section, I thought I could be more theoretical in my comments than in, say, the Help Me section.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Well, as you said, it depends on exactly what you can assume so Im making clear exactly what is assumed when dealing with real numbers; in this case, its the axioms for an ordered field.

I entirely agree.

Anyway, as this is in the Euler Avenue section, I thought I could be more theoretical in my comments than in, say, the Help Me section.

As much as possible, please.

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