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Context: Question made up by uni lecturer

Original statement: There exists two positive real numbers $x$ and $y$ such that for all positive integers $z$, $\frac{x}{y}>z$.

So the question was to find the negation of the statement, and then determine whether the original statement or its negation was true.

I found its negation to be: For all positive real numbers $x$ and $y$, there exists a positive integer $z$ such that $\frac{x}{y}\le z$.

The lecturer's solution to the question says that the negation is true since for any positive reals $x$ and $y$, you can choose $z$ to equal the ceiling of $\frac{x}{y}$.

When I attempted the question myself, I said that the original statement is true because you can take $x=z+1$ (which would be a positive integer that still belongs to the set of all positive real numbers) and $y=1$ (which is a positive real number), as $\frac{x}{y}=\frac{z+1}{1}=z+1>z$.

Can someone please help me to see the error in my answer.

Thanks

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The problem with your reasoning is the order. There exist $x,y$ positive integers such that for all positive integers $z$ we have $\frac{x}{y}>z$, so first you must pick an $x$ and a $y$, and then you must test whether $\frac{x}{y}>z$ for every positive integer $z$. Therefore you can't define $x=z+1$ as when you pick $x$ you don't know $z$ yet.

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"There exists two positive real numbers $x$ and $y$ such that for all positive integer $z$ : $\dfrac x y > z$"

The intuition says that it is False. $\dfrac x y$ is a positive real; thus, the statement amounts to asserting that there is a real that is greater than every integer, which is not.

You reasoning is wrong because you have swapped the choice of the numbers : you start from $z$ and choose $x$ and $y$ accordingly.

The negation of the original statement is : $\forall x \ \forall y \ \exists z \ (\dfrac y y \le z)$.

Thus, choose $x$ and $y$ positive whatever and what you get is a new positive real $\dfrac x y$.

Now, you have to choose an integer $z$ (obviously positive) that is greater-or-equal to $\dfrac x y$.

And this must be always possible, because $\dfrac x y$ is a number $r.r_1 r_2 r_3 \ldots$.

Consider as $z$ the number $r+1$ and it's done.

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