Evaluation of proofs This type of exercise will appear frequently in the book.
In each case, there is a proposed proof of a proposition. However, the
proposition may be true or may be false.
\- If a proposition is false, the proposed proof is, of course, incorrect. In
this situation, you are to find the error in the proof and then provide a
counterexample showing that the proposition is false.
\- If a proposition is true, the proposed proof may still be incorrect. In
this case, you are to determine why the proof is incorrect and then write a
correct proof using the writing guidelines that have been presented in this
book.
\- If a proposition is true and the proof is correct, you are to decide if the
proof is well written or not. If it is well written, then you simply must
indicate that this is an excellent proof and needs no revision. On the other
hand, if the proof is not well written, then you must then revise the proof by
writing it according to the guidelines presented in this text.(a) Proposition.
If \(m\) is an even integer, then \((5 m+4)\) is an even integer.
\mathrm{\\{} \text { Proof. We see that } 5 m + 4 = 1 0 n + 4 \(=2(5 n+2)\).
Therefore, \((5 m+4)\) is an even integer.
(b) Proposition. For all real numbers \(x\) and \(y,\) if \(x \neq y, x>0,\) and
\(y>0,\) then \(\frac{x}{y}+\frac{y}{x}>2\)
\mathrm{\\{} P r o o f . ~ S i n c e ~ \(x\) and \(y\) are positive real numbers,
\(x y\) is positive and we can multiply both sides of the inequality by \(x y\) to
obtain
$$
\begin{aligned}
\left(\frac{x}{y}+\frac{y}{x}\right) \cdot x y &>2 \cdot x y \\
x^{2}+y^{2} &>2 x y
\end{aligned}
$$
By combining all terms on the left side of the inequality, we see that
\(x^{2}-2 x y+y^{2}>0\) and then by factoring the left side, we obtain
\((x-y)^{2}>0 .\) Since \(x \neq y,(x-y) \neq 0\) and so \((x-y)^{2}>0 .\) This
proves that if \(x \neq y, x>0,\) and \(y>0,\) then \(\frac{x}{y}+\frac{y}{x}>2\)
(c) Proposition. For all integers \(a, b,\) and \(c,\) if \(a \mid(b c),\) then \(a
\mid b\) or \(a \mid c\).
\mathrm{\\{} P r o o f . ~ W e ~ a s s u m e ~ t h a t ~ \(a, b,\) and \(c\) are
integers and that \(a\) divides \(b c\). So, there exists an integer \(k\) such that
\(b c=k a\). We now factor \(k\) as \(k=m n,\) where \(m\) and \(n\) are integers. We
then see that
$$
b c=m n a .
$$
This means that \(b=m a\) or \(c=n a\) and hence, \(a \mid b\) or \(a \mid c\).
(d) Proposition. For all positive integers \(a, b,\) and
\(c,\left(a^{b}\right)^{c}=a^{\left(b^{c}\right)}\). This proposition is false
as is shown by the following counterexample:
If we let \(a=2, b=3,\) and \(c=2,\) then
$$
\begin{aligned}
\left(a^{b}\right)^{c} &=a^{\left(b^{c}\right)} \\
\left(2^{3}\right)^{2} &=2^{\left(3^{2}\right)} \\
8^{2} &=2^{9} \\
64 & \neq 512
\end{aligned}
$$
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