Inverse Inequality proof

jstarks4444 · **Joined:** Mon, 1 Nov 2010 17:13:40 UTC **Posts:** 71

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!

Shadow · **Posted:** Sat, 14 Jan 2012 22:21:32 UTC

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!

$a<b\iff b-a>0$ , in your case this means...

mathematic · **Joined:** Fri, 1 Jul 2011 01:17:26 UTC **Posts:** 322

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.

Shadow · **Posted:** Sun, 15 Jan 2012 04:16:01 UTC

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 ${1\over x}, {1\over y}>0$ ?

outermeasure · **Posted:** Sun, 15 Jan 2012 23:50:12 UTC

http://www.mathhelpforum.com/math-help/ ... 95287.html

mathematic · **Joined:** Fri, 1 Jul 2011 01:17:26 UTC **Posts:** 322

Shadow 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 ${1\over x}, {1\over y}>0$ ?

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)?

outermeasure · **Posted:** Mon, 16 Jan 2012 00:43:16 UTC

mathematic wrote:

Shadow 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 ${1\over x}, {1\over y}>0$ ?

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.

Shadow · **Posted:** Mon, 16 Jan 2012 09:35:41 UTC

Thank you outermeasure for elaborating in my absence. I recommended my approach because it's easier to show from what you already know.

outermeasure · **Posted:** Mon, 16 Jan 2012 10:09:58 UTC

Shadow wrote:

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 $\mathbb{R}$ .

mathematic · **Joined:** Fri, 1 Jul 2011 01:17:26 UTC **Posts:** 322

outermeasure wrote:

mathematic wrote:

Shadow 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 ${1\over x}, {1\over y}>0$ ?

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.

jstarks4444 · **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??

mathematic · **Joined:** Fri, 1 Jul 2011 01:17:26 UTC **Posts:** 322

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.

Shadow · **Posted:** Fri, 20 Jan 2012 04:15:41 UTC

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 ${1\over x}, {1\over y}>0$ when x,y>0

, in this way you can rename ${1\over xy}=a>0$ 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 ${1\over xy}$ and explain why it is positive, or perhaps wants him to explain the step where he distributes the multiplication by

over the difference y-x

.

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.

jstarks4444 · **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..

Shadow · **Posted:** Fri, 20 Jan 2012 18:31:22 UTC

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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Inverse Inequality proof

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