I need help with the following: Prove that for any $y$ in $\mathbb{R}$, there exists an $x$ in $\mathbb{R}$ such that $x-7>3y$. I tried to approach it by contradiction, leaving “there exists a $y$ such that for any $x$ such that $x-7\leq 3y$, though that didn’t seem very helpful either.
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Rewrite it as $x > 3y+7$. – Mauro ALLEGRANZA Oct 21 '18 at 17:20
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For any $y$, $3y+7$ is a number. Consider $(3y+7)+1$. – Mauro ALLEGRANZA Oct 21 '18 at 17:21
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The contradiction approach works too. Given an existing $y$, create a counterexample by picking $x=3y+8$. – Cuspy Code Oct 21 '18 at 17:31
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Note that $$x-7>3y \iff x>3y+7$$
Let $y$ be an arbitrary real number, the choose $x= 3y+8$ and we have $ x=3y+8 >3y+7$
Thus for every real number $y$ there exists an $x$ such as
$x=3y+8$ which satisfies $$ x-7>3y$$
Mohammad Riazi-Kermani
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Let $y\in \mathbb R$ then $3y+7 \in \mathbb R$
Hence now by Archimedean property, there exist $x \in \mathbb N (\subset \mathbb R$) such that
$x > 3y+7$
Which gives existence of required $x$.
Mayuresh L
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