Negation and the truth of propositions

Task number: 2783

First form the negation of each of the following propositions. Then decide whether the original proposition or its negation is true.

Variant 1
\(\forall x,y\in \mathbb R: x^2+y^2>0\)
Resolution
The negation of the proposition: \(\exists x,y\in R: x^2+y^2 \le 0\).

The negation is true, namely for \(x=y=0\).
Variant 2
\(\forall x\in \mathbb R\ \exists n\in \mathbb N: x< n\)
Resolution
Step by step: \( \neg (\forall x\in \mathbb R\ \exists n\in \mathbb N: x< n) \iff \\ \exists x\in \mathbb R: \neg (\exists n\in \mathbb N: x< n) \iff \\ \exists x\in \mathbb R\ \forall n\in \mathbb N: \neg(x<n) \)

So the negation is: \(\exists x\in \mathbb R\ \forall n\in \mathbb N: x\ge n\)

The original proposition is true, because by choosing \(n=\lceil |x|\rceil +1\) we have \(n\in \mathbb N\) and \(x \le |x| \le \lceil |x|\rceil <\lceil |x|\rceil +1=n\).
Variant 3
\(\forall x\in \mathbb R\ \exists n\in \mathbb N: (x\ge n) \land (x<n+1)\)
Resolution
The negation: \(\exists x\in \mathbb R\ \forall n\in \mathbb N: (x< n) \lor (x\ge n+1)\)

The negation is true; for example we may choose \(x=0\).
Variant 4
\(\forall \varepsilon> 0\ \exists \delta>0\ \forall x\in \mathbb R: |x-2|<\delta \ \Rightarrow\ |x-3| < \varepsilon \)
Resolution
The negation is \(\exists \varepsilon> 0\ \forall \delta>0\ \exists x\in \mathbb R: (|x-2|< \delta) \land (|x-3| \ge \varepsilon) \).

The negation is true. We choose \(\varepsilon =\frac12\) and irrespective of \(\delta\) we choose \(x=2\) . Then \(|x-2|=0< \delta\) is true for any \(\delta\) and also \(|x-3|=1\ge \frac12\) given our choice.
Variant 5
\(\forall \varepsilon> 0\ \exists \delta\ge 0\ \forall x\in \mathbb R: |x-2|<\delta \ \Rightarrow\ |x-3| < \varepsilon \)
(this differs from the previous variant in the range of \(\delta\))
Resolution
Negation: \(\exists \varepsilon> 0\ \forall \delta\ge 0\ \exists x\in R: (|x-2|< \delta) \land (|x-3| \ge \epsilon) \).

The original proposition is true. Irrespective of \(\varepsilon\) we choose \(\delta=0\). Because \(|x-2|<0\) is not true for any \(x\in \mathbb R\), the implication is always satisfied.
Variant 6
\(\forall x\in \mathbb N\ \exists y\in \mathbb N\ \forall z\in \mathbb N: z>x \Rightarrow y<z\)
Resolution
The negation is \(\exists x\in \mathbb N\ \forall y\in \mathbb N\ \exists z\in \mathbb N: z>x \land y\ge z\)

The original proposition is true; we choose \(y=x\), because given this choice the implication \(z>x \Rightarrow y<z\) is equivalent to \(z>x \Rightarrow x<z\) and is always satisfied.
Variant 7
\(\exists y\in \mathbb N\ \forall x\in \mathbb N\ \forall z\in \mathbb N: z>x \Rightarrow y<z\)
Resolution
The negation is \(\forall y\in \mathbb N\ \exists x\in \mathbb N\ \exists z\in \mathbb Z: z>x \land y\ge z\)

The original proposition is true; we choose \(y=1\). From the assumption \(z>x\), \(x,z\in \mathbb N\) we get \(z\ge 2\). So the consequence \(y=1<z\) is always true.
Variant 8
\(\exists y\in \mathbb N\ \forall x\in \mathbb N\ \forall z\in \mathbb R: z>x \Rightarrow y<z\)
Resolution
The negation is \(\forall y\in \mathbb N\ \exists x\in \mathbb N\ \exists z\in \mathbb R: z>x \land y\ge z\)

The original proposition is true; we choose \(y=1\). From the assumption \(z>x\), \(x\in \mathbb N, z\in \mathbb R\) we get \(z> 1\). So the consequence \(y=1<z\) is always true.
Variant 9
\(\exists y\in \mathbb N\ \forall x\in \mathbb R\ \forall z\in \mathbb N: z>x \Rightarrow y<z\)
Resolution
The negation is \(\forall y\in \mathbb N\ \exists x\in \mathbb R\ \exists z\in \mathbb N: z>x \land y\ge z\)

The negation is true; we choose \(x=0\) and \(z=y\). Then \(z>x\) and also \(y\le z\).