Solutions to More Practice on Proofs
Solutions & Commentary to Accompany More Practice on Proofs
The questions presented here have been adapted from Kenneth Rosen’s Discrete Mathematics and Its Applications, 8th Edition and Oscar Levin’s Discrete Mathematics:An Open Introduction.
Problem 1 ¶
Use a direct proof to show that the sum of two odd integers is even.
Let $n = 2k + 1$ and $m = 2l + 1$ be odd integers. Then $n+m=2k+1+2l+1=2k+2l+2=2(k+l+1)$To make this more explicit, we can write $2(k+l+1)$ as $2(\alpha)$, where $\alpha = k+l+1$ is even.
Problem 2 ¶
Show that if $n$ is an integer and $n^3 + 5$ is odd, then $n$ is even
Proof by Contraposition ¶
Assume that $n$ is odd, so $n = 2k + 1$ for some integer $k$. Then $$n^3 +5 = (2k+1)^3+5=(8k^3+12k^2+6k+1)+5$$$$=8k^3+12k^2+6k+6= 2(4k^3 +6k^2 +3k +3).$$ Because $n^3 + 5$ is two times some integer, it is even.
Proof by Contradiction ¶
Suppose that $n^3 + 5$ is odd and $n$ is odd. Because $n$ is odd and the product of two odd numbers is odd, it follows that $n^2$ is odd and then that $n^3$ is odd. But then $5 = (n^3 + 5) − n^3$ would have to be even because it is the difference of two odd numbers. Therefore, the supposition that $n^3 + 5$ and $n$ were both odd is wrong.
Problem 3 ¶
Prove that if $n$ is an odd integer, then $n^3$ is an odd integer
$n$ is odd, therefore $n = 2k + 1$ for some integer $k$. Then $$n^3 = (2k+1)^3=(8k^3+12k^2+6k+1)$$$$=8k^3+12k^2+6k+1= 2(4k^3 +6k^2 +3k +1)+1.$$ Since $n^3$ is 1 plus 2 times some integer it is odd.To make this more explicit, we can write $2(4k^3 +6k^2 +3k +1)+1$ as $2(\alpha)+1$, where $\alpha = 4k^3 +6k^2 +3k +1$
Problem 4 ¶
Prove that if $x^3$ is irrational, then $x$ is irrational.
Let’s prove the contrapositive: If $x$ is rational, then $x^3$ is also rational.If a number is not irrataional, it is rational
Since $x$ is rational, we can write $x=\frac{p}{q}$ where $p,q$ are integers with $q \neq 0$.$q \neq 0$ is an important note to make, since division by 0 is undefined. $\implies x^3 = (\frac{p}{q})^3=\frac{p^3}{q^3}$ which is irrational (because $p^3$ and $q^3$ are integers).
Therefore by contrapositive, we have show that:if $x^3$ is irrational, then $x$ is irrational.
Problem 5 ¶
Prove that for all integers $a$ and $b$, if $a + b$ is odd, then $a$ is odd or $b$ is odd.
The problem with trying a direct proof is that it will be hard to separate $a$ and $b$ from knowing something about $a + b$. On the other hand, if we know something about $a$ and $b$ separately, then combining them might give us information about $a+b$. The contrapositive of the statement we are trying to prove is: for all integers $a$ and $b$, if $a$ and $b$ are even, then $a+b$ is even. Note that our assumption that $a$ and $b$ are even is really the negation of $a$ is odd or $b$ is odd. We used De Morgan’s law here.
Thus our proof will have the following format:
- Let $a$ and $b$ be integers. Assume that $a$ and $b$ are both even.
- Then, $a=2k$ and and $b =2l$ for some integers $k$and $l$.
- Now, $a+b=2k+2l=2(k+l)$
- Since $k+l$ is an integer (let’s called it $\alpha$), $a+b=2(k+l)=2(\alpha)$ is an even integer $\blacksquare$