# Application of Biconditional

• Oct 24th 2010, 06:31 AM
novice
Application of Biconditional
Prove that $\displaystyle \forall x \in \mathbb{R}[\exists y \in \mathbb{R}(x+y=xy)\Leftrightarrow x \not = 1]$

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

Let $\displaystyle Q:x \not = 1$

For $\displaystyle P\Rightarrow Q$, I did this:

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

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

Now $\displaystyle Q \Rightarrow P$:

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

I am cheating here:

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

Does any one have a convincing proof for this part?
• Oct 24th 2010, 07:37 AM
tonio
Quote:

Originally Posted by novice
Prove that $\displaystyle \forall x \in \mathbb{R}[\exists y \in \mathbb{R}(x+y=xy)\Leftrightarrow x \not = 1]$

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

Let $\displaystyle Q:x \not = 1$

For $\displaystyle P\Rightarrow Q$, I did this:

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

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

Now $\displaystyle Q \Rightarrow P$:

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

I am cheating here:

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

Does any one have a convincing proof for this part?

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

Tonio
• Oct 24th 2010, 12:46 PM
novice
Quote:

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

Tonio

No doubt it's solvable, but how do you prove the converse?
• Oct 24th 2010, 08:27 PM
tonio
Quote:

Originally Posted by novice
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