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 Post subject: Inverse Inequality proofPosted: Sat, 14 Jan 2012 22:19:43 UTC
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Joined: Mon, 1 Nov 2010 17:13:40 UTC
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Hey all, would really like some help with the following proof:

Let x,y ∈ R>0 (real numbers greater than 0). If x < y then 0 < 1/y < 1/x

All help appreciated, thanks a bunch!

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 Post subject: Re: Inverse Inequality proofPosted: Sat, 14 Jan 2012 22:21:32 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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jstarks4444 wrote:
Hey all, would really like some help with the following proof:

Let x,y ∈ R>0 (real numbers greater than 0). If x < y then 0 < 1/y < 1/x

All help appreciated, thanks a bunch!

, in your case this means...

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 Post subject: Re: Inverse Inequality proofPosted: Sun, 15 Jan 2012 04:06:17 UTC
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jstarks4444 wrote:
Hey all, would really like some help with the following proof:

Let x,y ∈ R>0 (real numbers greater than 0). If x < y then 0 < 1/y < 1/x

All help appreciated, thanks a bunch!

(I also answered in another forum).

0 < x < y => 0 < x/y < 1 => 0 < 1/y < 1/x. The point being in both steps we are dividing by a positive number, so the direction of the inequality doesn't change.

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 Post subject: Re: Inverse Inequality proofPosted: Sun, 15 Jan 2012 04:16:01 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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mathematic wrote:
jstarks4444 wrote:
Hey all, would really like some help with the following proof:

Let x,y ∈ R>0 (real numbers greater than 0). If x < y then 0 < 1/y < 1/x

All help appreciated, thanks a bunch!

(I also answered in another forum).

0 < x < y => 0 < x/y < 1 => 0 < 1/y < 1/x. The point being in both steps we are dividing by a positive number, so the direction of the inequality doesn't change.

How do you know that ?

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 Post subject: Re: Inverse Inequality proofPosted: Sun, 15 Jan 2012 23:50:12 UTC
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 Post subject: Re: Inverse Inequality proofPosted: Mon, 16 Jan 2012 00:31:27 UTC
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mathematic wrote:
jstarks4444 wrote:
Hey all, would really like some help with the following proof:

Let x,y ∈ R>0 (real numbers greater than 0). If x < y then 0 < 1/y < 1/x

All help appreciated, thanks a bunch!

(I also answered in another forum).

0 < x < y => 0 < x/y < 1 => 0 < 1/y < 1/x. The point being in both steps we are dividing by a positive number, so the direction of the inequality doesn't change.

How do you know that ?

Did you understand the first step (divide by y does not change direction of inequality)? Did you understand the second step (divide by x does not change the direction of the inequality)?

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 Post subject: Re: Inverse Inequality proofPosted: Mon, 16 Jan 2012 00:43:16 UTC
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mathematic wrote:
mathematic wrote:
jstarks4444 wrote:
Hey all, would really like some help with the following proof:

Let x,y ∈ R>0 (real numbers greater than 0). If x < y then 0 < 1/y < 1/x

All help appreciated, thanks a bunch!

(I also answered in another forum).

0 < x < y => 0 < x/y < 1 => 0 < 1/y < 1/x. The point being in both steps we are dividing by a positive number, so the direction of the inequality doesn't change.

How do you know that ?

Did you understand the first step (divide by y does not change direction of inequality)? Did you understand the second step (divide by x does not change the direction of the inequality)?

"Divide by y (or x) does not change direction of inequality": you need to prove that from the ~12 defining axioms of ordered field. If you do it properly, you will see where Shadow is coming from.

Note that we don't need the completeness axiom for this question.

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 Post subject: Re: Inverse Inequality proofPosted: Mon, 16 Jan 2012 09:35:41 UTC
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Thank you outermeasure for elaborating in my absence. I recommended my approach because it's easier to show from what you already know.

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 Post subject: Re: Inverse Inequality proofPosted: Mon, 16 Jan 2012 10:09:58 UTC
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Thank you outermeasure for elaborating in my absence. I recommended my approach because it's easier to show from what you already know.

Me too. Given this question is A&NT and not Algebra subforum, the expectation should be an axiom-grubbing exercise from the defining properties of .

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 Post subject: Re: Inverse Inequality proofPosted: Tue, 17 Jan 2012 00:51:23 UTC
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outermeasure wrote:
mathematic wrote:
mathematic wrote:
jstarks4444 wrote:
Hey all, would really like some help with the following proof:

Let x,y ∈ R>0 (real numbers greater than 0). If x < y then 0 < 1/y < 1/x

All help appreciated, thanks a bunch!

(I also answered in another forum).

0 < x < y => 0 < x/y < 1 => 0 < 1/y < 1/x. The point being in both steps we are dividing by a positive number, so the direction of the inequality doesn't change.

How do you know that ?

Did you understand the first step (divide by y does not change direction of inequality)? Did you understand the second step (divide by x does not change the direction of the inequality)?

"Divide by y (or x) does not change direction of inequality": you need to prove that from the ~12 defining axioms of ordered field. If you do it properly, you will see where Shadow is coming from.

Note that we don't need the completeness axiom for this question.

I will plead ignorance. I didn't know that it was necessary to prove the obvious, but I guess it has to be for this question.

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 Post subject: Re: Inverse Inequality proofPosted: Fri, 20 Jan 2012 00:23:37 UTC
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Joined: Mon, 1 Nov 2010 17:13:40 UTC
Posts: 71
Yeah haha, we have to be more axiomatic in our approach..

Now, I have the following proof set up but I was told there is a minor mistake in it:

x < y means y-x ∈ N by definition of inequality
So y-x > 0 by definition of natural numbers
y/(xy) - x/(xy) > 0/(xy)
1/x - 1/y > 0
1/x - 1/y > 0 means 1/y - 1/x by definition of inequality
0 < 1/y < 1/y by a proposition which states x ∈ R>0 (positive real numbers) if and only if 1/x ∈ R>0 (positive real numbers)

Can anyone tell me what is wrong here??

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 Post subject: Re: Inverse Inequality proofPosted: Fri, 20 Jan 2012 00:43:41 UTC
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y/(xy) - x/(xy) > 0/(xy)

The above statement (which is obviously true) suffers from the same flaw that the proof I presented does, i.e. the inequality is preserved when dividing by a positive constant.

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 Post subject: Re: Inverse Inequality proofPosted: Fri, 20 Jan 2012 04:15:41 UTC
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mathematic wrote:
y/(xy) - x/(xy) > 0/(xy)

The above statement (which is obviously true) suffers from the same flaw that the proof I presented does, i.e. the inequality is preserved when dividing by a positive constant.

No it does not, this other one only requires that you know that when , in this way you can rename and use the original suggestion I gave, the problem with your suggestion is that you are assuming something about the order relations which you haven't proven, but just knowing positivity is weaker. He has a proposition which tells him that this second part is true. I don't see the error in the proof, unless the grader wants him to specifically state the substitution of for and explain why it is positive, or perhaps wants him to explain the step where he distributes the multiplication by over the difference .

Also, to the op: I think you have a typo on line 5 and you want an inequality instead of another minus sign after the word "means", and line 6 doesn't make sense at all the way you've written it.

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 Post subject: Re: Inverse Inequality proofPosted: Fri, 20 Jan 2012 18:30:42 UTC
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Joined: Mon, 1 Nov 2010 17:13:40 UTC
Posts: 71
Ah yes my bad that - is meant to be a <

For line 6, I am saying that since we know x,y ∈ R>0 , then we can say that 1/x and 1/y are also ∈ R>0

I was told that the steps are correct but there is some reasoning that I have not spelt out..

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 Post subject: Re: Inverse Inequality proofPosted: Fri, 20 Jan 2012 18:31:22 UTC
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jstarks4444 wrote:
Ah yes my bad that - is meant to be a <

For line 6, I am saying that since we know x,y ∈ R>0 , then we can say that 1/x and 1/y are also ∈ R>0

I was told that the steps are correct but there is some reasoning that I have not spelt out..

Oh, then it's probably the things I mentioned in my last post.

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