1. ## 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?

2. Originally Posted by novice
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

3. Originally Posted by tonio
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?

4. 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