Results 1 to 4 of 4

Math Help - Application of Biconditional

  1. #1
    Banned
    Joined
    Sep 2009
    Posts
    502

    Application of Biconditional

    Prove that \forall x \in \mathbb{R}[\exists y \in \mathbb{R}(x+y=xy)\Leftrightarrow x \not = 1]

    Let P:\forall x \in \mathbb{R}[\exists y \in \mathbb{R}(x+y=xy).



    Let Q:x \not = 1

    For P\Rightarrow Q, I did this:

    For \forall x \in \mathbb{R} \exists y \in \mathbb{R},,

    x+y=xy \Rightarrow xy-y=x \Rightarrow y=\frac{x}{x-1}\Rightarrow x \not = 1

    Now Q \Rightarrow P:

    x\not = 1 means x >1 or x<1, but the possibility is endless.

    I am cheating here:

    x \not = 1 \Rightarrow x>1 or x<1
    Without lost of generality, say x>1. Then for some y, xy >y \Rightarrow xy=y+x. I know this is not true for all x.

    Does any one have a convincing proof for this part?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by novice View Post
    Prove that \forall x \in \mathbb{R}[\exists y \in \mathbb{R}(x+y=xy)\Leftrightarrow x \not = 1]

    Let P:\forall x \in \mathbb{R}[\exists y \in \mathbb{R}(x+y=xy).



    Let Q:x \not = 1

    For P\Rightarrow Q, I did this:

    For \forall x \in \mathbb{R} \exists y \in \mathbb{R},,

    x+y=xy \Rightarrow xy-y=x \Rightarrow y=\frac{x}{x-1}\Rightarrow x \not = 1

    Now Q \Rightarrow P:

    x\not = 1 means x >1 or x<1, but the possibility is endless.

    I am cheating here:

    x \not = 1 \Rightarrow x>1 or x<1
    Without lost of generality, say x>1. Then for some y, xy >y \Rightarrow xy=y+x. I know this is not true for all x.

    Does any one have a convincing proof for this part?

    For any x : x+y=xy\Longleftrightarrow x+y(1-x)=0\Longleftrightarrow y(1-x)=-x , and this last equation is solvable iff x\neq 1.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Sep 2009
    Posts
    502
    Quote Originally Posted by tonio View Post
    For any x : x+y=xy\Longleftrightarrow x+y(1-x)=0\Longleftrightarrow y(1-x)=-x , and this last equation is solvable iff x\neq 1.

    Tonio
    No doubt it's solvable, but how do you prove the converse?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by novice View Post
    No doubt it's solvable, but how do you prove the converse?

    What converse? Didn't you notice the double arrows? The proof above is the WHOLE biconditional proof!

    Tonio
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Biconditional Statement
    Posted in the Geometry Forum
    Replies: 2
    Last Post: December 21st 2008, 12:48 PM
  2. Replies: 1
    Last Post: October 9th 2008, 04:17 AM
  3. Replies: 1
    Last Post: July 3rd 2007, 03:59 PM
  4. application 3
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 18th 2007, 03:36 AM
  5. application 2
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 18th 2007, 03:30 AM

Search Tags


/mathhelpforum @mathhelpforum