Homework Answers
5(a). Given sets $S$ and $T$, then $(x \in S) \land (x \in T)
\Rightarrow x\in(S \cap T)$.
5(b). Given any two sets such that some element belongs to both these sets, then the element belongs to the intersection of the two sets.
Bonus problem: We will prove by contradiction. Let S= a+b, where S, a and b are some natural numbers such that S is an odd natural number.
case 1: Assume both a and are odd. Since a and b are odd, a and b can be expressed as follows. a=2m+1 and b=2n+1, where n and m are natural numbers greater than or equal to 0. So, S=a+b=2m+1+2n+1=2(m+n+1). This means, S is multiple of 2 and is therefore an even number. This contradicts the fact that S is an odd number. Therefore, by contradiction, both a and b cannot be odd numbers.
case2: Assume none of a and b are odd. Then, both a and b are
even numbers. So, we can express a and b as follows: a=2p and b=2q,
where p and q are natural numbers greater than or equal to 0. Then,
S=a+b=2p+2q=2(p+q). This means, S is a multiple of 2 and is
therefore an even number. This contradicts the fact that S is an
odd number. Therefore, by contradiction, both a and b cannot be
even numbers.
Since a and b cannot be both odd(case1) and cannot be both even (case 2), the only possiblilty is that exactly one of them is odd. Hence proved.
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