8. In the previous problem, put a negation in front of the logical expression for “Someone in your class is perfect”, then move the negation until negation only appears directly in front of S(x) or P(x), by applying DeMorgan’s Laws. 9. Let these be the hypotheses: If it’s not cold or if it’s not windy, then I will go walking. If I go walking, I’ll feel good. I don’t feel good. Use rules of inference to show that the above hypotheses imply the conlusion:“It’s windy.” 10. Use a proof by contraposition to prove that if m and n are integers and mn is even, then m is even or n is even. 11. Use a proof by contradiction to prove that if x is an irrational number and y is a rational number, then x + y is an irrational number.

10.

Suppose m and n are odd. then:

m=2k+1, n=2l+1

$mn=(2k+1)(2l+1)=4kl+2k+2l+1$ is odd number.

So, m is even or n is even.

11.

We have:

$x\in Q$ (the set of rational numbers)

$y\in \{R/Q\}$ (the set of irrational numbers)

Let us assume that $x+y\in Q$ , ie $x+y$ is rational, then:

$\exist m,n,p,q\in Z$ st $x=m/n$ (since x is rational ), and $x+y=p/q$ (since the sum is rational).

Therefore, we can write;

$m/n+y=p/q$

$y=\frac{p}{q}-\frac{m}{n}=\frac{np-mq}{nq}$

And so y can be written as a fraction$\implies$ y is rational.

But we initially asserted that y was irrational and hence we have a contradiction, and so the sum

x+y cannot be rational and hence it must be irrational.

8.

S(x): "in your class"

P(x): "is perfect"

$\exist xS(x)P(x):$ "Someone in your class is perfect"

by DeMorgan’s Laws:

$\neg (\exist xS(x)P(x))\equiv \forall x \neg (S(x)P(x))\equiv \forall x (\neg S(x)\lor \neg P(x))$ :

"Everybody are in class, or everybody are perfect ".

9.

I don’t feel good$\implies$ I do not go walking

I do not go walking$\implies$ it’s cold or it’s windy