8.14 Quantifiers Tutorial

Prove that for all \(x\in\mathbb{R}\) if x\(^2\) is irrational. then x is irrational
Suppose that \(x\) is rational, then \(x=\frac{m}{n}\) is rational and \(n\neq0\)
Hence \(x^2=(\frac{\mathrm{m}}{\mathrm{n}})^2=\frac{\mathrm{m}^2}{\mathrm{n}^2}\), with \(\mathrm{m^2,n^2\in\mathbb{Z}}\) and \(n^2 \neq0\)
Thus \(x^2\) is rational, since \(x^2\) is irrational.

Contradiction!
We have that for all \(x\in\mathbb{R}\), if \(x^2\) is irrational,then \(x\) is irrational.
Q: \(p(m)=" m^{2}\) is even"\(m \in N\), \(q(m)=" m\) is even" \(m \in N\). Prove that \(\mathrm{p}(\mathrm{m}) \Rightarrow \mathrm{q}(\mathrm{m})\)
To prove that \(P \Rightarrow Q\) using contradiction we need to prove that \((P\wedge \neg Q)\Rightarrow(R\wedge \neg R)\)
Suppose that \(n^2\) is even(\(R\)) and \(n\) is odd.
Since \(n\) is odd then \(n~=~2k+1\) for some \(k \in\mathbb{Z}\)
Hence \(n^2 = (2k+1)^2 = 4k^2+4k+1 = 2(2k^2+2k)+1\) is odd (\(\neg R\))which is a contradiction.
Prove\(~2^n+1\) is a prime for all positive integer number \(n\)

Let try some examples.

\(n=1\), \(2^n+1=3~yes\). \(n=2\), \(2^n+1=5~yes\). \(n=3\), \(2^n+1=9~No\)
Prove \((\forall x)(p(x)\vee q(x))\Leftarrow(\forall x)(p(x))\vee(\forall x)(q(x))\)

By contrapositive, the question is equivalent to prove

\(\neg[(\forall x)(p(x)\lor q(x))]\Rightarrow\neg[(\forall x)(p(x))\lor(\forall x)(q(x))]\)

\((\exists x)(\neg p(x)\wedge \neg q(x))\Rightarrow \neg(\exists x)(\neg p(x))\wedge (\exists x)(\neg p(x))\)

So exist \(x_0\), such that \(\neg p(x_0)\) and \(\neg q(x_0)\) are both true

Hence it means that exist \(x_0\), such that\(\neg p(x_0)\) is true and exist \(x_0\), such that\(\neg q(x_0)\) is true

Therefore it is equivalent to \((\exists x)(\neg p(x))\wedge (\exists x)(\neg q(x))\)

Therefore \((\exists x)(\neg p(x)\wedge \neg q(x))\Rightarrow \neg(\exists x)(\neg p(x))\wedge (\exists x)(\neg p(x))\)

By contrapositive, we can get \((\forall x)(p(x)\vee q(x))\Leftarrow(\forall x)(p(x))\vee(\forall x)(q(x))\)

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