MAT102 Lecture 5

• $\mathrm{\forall }n\in \mathbb{Z},\mathrm{\forall }m\in \mathbb{Z},\mathrm{\exists }r\in \mathbb{Q},[(n<m)⟹(n<r)\wedge (r<m)]$
• This is the negation.
• $\mathrm{\exists }n\in \mathbb{Z},\mathrm{\exists }m\in \mathbb{Z},\mathrm{\forall }R\in \mathbb{Q},(r\le n<m)\vee (n<m\le r)$

1

A

• Prove $P⟹(Q\vee R)$
• Let's setup a truth table to show that $P⟹(Q\vee R)$

• We can see that $\mathrm{\neg }P\vee (Q\vee R)=(\mathrm{\neg }P\vee Q)\vee R$ with the truth table constructed above.

B

• $D(x)=\text{ "Is not divisible by two"}$
• $m,n\in \mathbb{Z},(D(m^3+n^3))⟹D(m)\vee D(n)$
• $P=m,n\in \mathbb{Z},D(m^3+n^3)$
• $Q=D(m)$
• $R=D(n)$
• $P⟹(Q\vee R)$
• $(P\wedge \mathrm{\neg }Q)⟹R$
• $m,n\in \mathbb{Z},(D(m^3+n^3))\wedge \mathrm{\neg }D(m)⟹D(n)$
• By part A, we know that $P⟹(Q\vee R)$ is the same as $(P\wedge \mathrm{\neg }Q)⟹R$. So this means that if $m^3+n^3$ is odd and $m$ is even, that $n$ must be odd.
  • $(P\wedge \mathrm{\neg }Q)⟹R$
  • $R⟹(P\wedge \mathrm{\neg }Q)$
  • $\mathrm{\neg }R⟹\mathrm{\neg }(P\wedge \mathrm{\neg }Q)$
  • $\mathrm{\neg }R⟹\mathrm{\neg }P\vee Q$
• Assume that $\mathrm{\neg }D(m)$ or m is even.
• $k\in \mathbb{Z}$
• $m=2k$
• $(2k{)}^3+n^3$
• $8k^3+n^3$
• This must be odd per our statement. Since $8k^3$ is even, this must mean that $n^3$ is odd. Because if $n^3$ is even, the sum of two even numbers would be even.

• Suppose that $m^3+n^3$ is odd, and $m$ is even.
• Claim: If $m$ is even then $m^3$ is even.
• If $m=2k$ for $k\in \mathbb{Z}$, then $m^3=8k^3=2(4k^3)$ is even.
• $n^3=(m^3+n^3)-m^3$
  • Odd-even=odd
• Now we know $n^3$ is odd. We want to show that $n$ is odd.
• By the contrapositive of the first claim. $n$ must be odd.