S.O.S. Mathematics CyberBoard

Your Resource for mathematics help on the web!
It is currently Thu, 23 May 2013 18:36:09 UTC

All times are UTC [ DST ]




Post new topic Reply to topic  [ 16 posts ]  Go to page 1, 2  Next
Author Message
 Post subject: Inverse Inequality proof
PostPosted: Sat, 14 Jan 2012 22:19:43 UTC 
Offline
Senior Member

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!


Top
 Profile  
 
 Post subject: Re: Inverse Inequality proof
PostPosted: Sat, 14 Jan 2012 22:21:32 UTC 
Offline
Moderator
User avatar

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
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...

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination


Top
 Profile  
 
 Post subject: Re: Inverse Inequality proof
PostPosted: Sun, 15 Jan 2012 04:06:17 UTC 
Offline
Member of the 'S.O.S. Math' Hall of Fame

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.


Top
 Profile  
 
 Post subject: Re: Inverse Inequality proof
PostPosted: Sun, 15 Jan 2012 04:16:01 UTC 
Offline
Moderator
User avatar

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
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?

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination


Top
 Profile  
 
 Post subject: Re: Inverse Inequality proof
PostPosted: Sun, 15 Jan 2012 23:50:12 UTC 
Online
Moderator
User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6007
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
http://www.mathhelpforum.com/math-help/ ... 95287.html

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


Top
 Profile  
 
 Post subject: Re: Inverse Inequality proof
PostPosted: Mon, 16 Jan 2012 00:31:27 UTC 
Offline
Member of the 'S.O.S. Math' Hall of Fame

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)?


Top
 Profile  
 
 Post subject: Re: Inverse Inequality proof
PostPosted: Mon, 16 Jan 2012 00:43:16 UTC 
Online
Moderator
User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6007
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
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.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


Top
 Profile  
 
 Post subject: Re: Inverse Inequality proof
PostPosted: Mon, 16 Jan 2012 09:35:41 UTC 
Offline
Moderator
User avatar

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
Thank you outermeasure for elaborating in my absence. I recommended my approach because it's easier to show from what you already know.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination


Top
 Profile  
 
 Post subject: Re: Inverse Inequality proof
PostPosted: Mon, 16 Jan 2012 10:09:58 UTC 
Online
Moderator
User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6007
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
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}.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


Top
 Profile  
 
 Post subject: Re: Inverse Inequality proof
PostPosted: Tue, 17 Jan 2012 00:51:23 UTC 
Offline
Member of the 'S.O.S. Math' Hall of Fame

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.


Top
 Profile  
 
 Post subject: Re: Inverse Inequality proof
PostPosted: Fri, 20 Jan 2012 00:23:37 UTC 
Offline
Senior Member

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??


Top
 Profile  
 
 Post subject: Re: Inverse Inequality proof
PostPosted: Fri, 20 Jan 2012 00:43:41 UTC 
Offline
Member of the 'S.O.S. Math' Hall of Fame

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.


Top
 Profile  
 
 Post subject: Re: Inverse Inequality proof
PostPosted: Fri, 20 Jan 2012 04:15:41 UTC 
Offline
Moderator
User avatar

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
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 a for {1\over xy} and explain why it is positive, or perhaps wants him to explain the step where he distributes the multiplication by a 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.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination


Top
 Profile  
 
 Post subject: Re: Inverse Inequality proof
PostPosted: Fri, 20 Jan 2012 18:30:42 UTC 
Offline
Senior Member

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..


Top
 Profile  
 
 Post subject: Re: Inverse Inequality proof
PostPosted: Fri, 20 Jan 2012 18:31:22 UTC 
Offline
Moderator
User avatar

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
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.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination


Top
 Profile  
 
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 16 posts ]  Go to page 1, 2  Next

All times are UTC [ DST ]


Who is online

Users browsing this forum: No registered users


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum

Search for:
Jump to:  
Contact Us | S.O.S. Mathematics Homepage
Privacy Statement | Search the "old" CyberBoard

users online during the last hour
Powered by phpBB © 2001, 2005-2011 phpBB Group.
Copyright © 1999-2013 MathMedics, LLC. All rights reserved.
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA