Velamma Episode 24 Pdf Free ^HOT^ 40

Velamma Episode 24 Pdf Free ^HOT^ 40

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Velamma Episode 24 Pdf Free 40

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Proving contradiction

“$F(x):\exists y(x \leq y) \land (y \leq x) \land (x,y \in \mathbb{R})$ and $F(x):\forall y(y\leq x)\land (x,y \in \mathbb{R})$ where F is a relation on $\mathbb{R}$.
We want to show that $F$ is transitive. That is, given $x,y \in \mathbb{R}$, if $x \leq y$ and $y \leq x$, then $x=y$. For this, let $x \leq y$ and $y \leq x$ and $x
eq y$ be given. Now, $x \leq y \implies F(x)$. Similarly, $F(y)$. But this implies the existence of some $y$ satisfying $x \leq y$ and $y \leq x$ since $F$ is transitive. But this is impossible since $x
eq y$. Therefore, we have a contradiction.
Is my logic sound? If not, please help!


Your logic is sound. You have made use of the fact that $F$ is transitive.
If $F$ were not transitive, there would be some $x \in \mathbb{R}$ such that $F(x)$ but $
eg F(y)$ for all $y \le x$. This implies that $
eg F(y)$ for all $y \in \mathbb{R}$. This is a contradiction, since the empty set is not a real number.
Thus, $F$ must be transitive.



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