1.
Suppose \(a\) and \(b\) are odd. That is, \(a=2k+1\) and \(b=2m+1\) for some integers \(k\) and \(m\text{.}\) Then
\begin{align*}
ab \amp =(2k+1)(2m+1)\\
\amp =4km+2k+2m+1\\
\amp =2(2km+k+m)+1\text{.}
\end{align*}
Therefore \(ab\) is odd.
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Consider the statement:
Which of the proofs below appear to be valid proofs of this statement? Note: You can assume all the algebra below is correct (because it is).
2.
Assume that \(a\) or \(b\) is even -- say it is \(a\) (the case where \(b\) is even will be identical). That is, \(a=2k\) for some integer \(k\text{.}\) Then
\begin{align*}
ab \amp =(2k)b\\
\amp =2(kb)\text{.}
\end{align*}
Thus \(ab\) is even.
3.
Suppose that \(ab\) is even but \(a\) and \(b\) are both odd. Namely, \(ab = 2n\text{,}\) \(a=2k+1\) and \(b=2j+1\) for some integers \(n\text{,}\) \(k\text{,}\) and \(j\text{.}\) Then
\begin{align*}
2n \amp =(2k+1)(2j+1)\\
2n \amp =4kj+2k+2j+1\\
n \amp = 2kj+k+j+\frac{1}{2}\text{.}
\end{align*}
But since \(2kj+k+j\) is an integer, this says that the integer \(n\) is equal to a non-integer, which is impossible.
4.
Let \(ab\) be an even number, say \(ab=2n\text{,}\) and \(a\) be an odd number, say \(a=2k+1\text{.}\)
\begin{align*}
ab \amp =(2k+1)b\\
2n \amp =2kb+b\\
2n-2kb\amp =b\\
2(n-kb)\amp =b\text{.}
\end{align*}
Therefore \(b\) must be even.
