是非題

• If $$x$$ and $$y$$ are integers of the same parity, then $$xy$$ and $$(x+y)^2$$ are of the same parity. (Two integers are of the same parity if they are both odd or both even.)

Solution.
Let $$x = y = 1$$. Then $$x$$ and $$y$$ are of the same parity. However, $$xy = 1$$ and $$(x+y)^2 = 4$$ are of distinct parities. $$\square$$

運算元混淆

• Let $$A$$ and $$B$$ be two sets. If $$A \setminus B = B \setminus A$$, then $$A \setminus B = \varnothing$$.

(錯誤寫法)
Since $$A \setminus B = A \setminus (A \cap B)$$ and $$B \setminus A = B \setminus (A \cap B)$$, we have \begin{aligned} A \setminus (A \cap B) = B \setminus (A \cap B) \quad \Rightarrow \quad A = B \quad \Rightarrow \quad A \setminus B = \varnothing\end{aligned}

Proof.
Suppose that $$A \setminus B \neq \varnothing$$. Let $$x \in A \setminus B$$. Then $$x \in A$$ but $$x \notin B$$. Since $$A \setminus B = B \setminus A$$, it is seen that $$x \in B$$ but $$x \notin A$$. This shows that $$x \in A$$ and $$x \notin A$$, which is a contradiction. Hence $$A \setminus B = \varnothing$$. $$\square$$

說明不夠嚴謹

• For every positive irrational number $$b$$, there is an irrational number $$a$$ such that $$0 < a < b$$.

倒因為果

• $$n^3 + 1 > n^2 + n$$ for every integer $$n \geq 2$$.

(錯誤寫法)
Suppose that $$n^3 + 1 > n^2 + n$$ for every integer $$n \geq 2$$. Then \begin{aligned} n^3 + 1 > n^2 + n \quad &\Rightarrow \quad n^3 + 1 - n^2 - n > 0 \\ &\Rightarrow \quad n^2(n-1) - (n-1) > 0 \\ &\Rightarrow \quad (n-1)(n^2-1) > 0 \\ &\Rightarrow \quad (n-1)^2 (n+1) > 0\end{aligned} The last inequality is always true for every integer $$n \geq 2$$. $$\square$$