# Proofs True and False Statements.

So I want to just check if my my critical reading of these problems are correct. I want to see if my answers match up.

(a) $\forall x\in\mathbb{Z}$: if $x^2\leqslant 30$, then $x\leqslant5$.

(b) $\forall x\in\mathbb{R}$: if $x^2\leqslant 30$, then $x\leqslant 5$.

(c) $\forall x\in\mathbb{Z}$: if $x\leqslant 5$, then $x^2\leqslant 30$.

(d) $\forall x\in\mathbb{R}_+$: if $x\leqslant 5$, then $x^2\leqslant 30$.

It seems like the first three statements are false and the last one is true but I could be mistaken. Thoughts?

• a) is not correct $-5\leq x\leq 5$, b) $\sqrt{-5}\leq x\leq\sqrt{5}$ c) what about $x=-8$ d) is true since $x^2$ is an increasing function on $[0, \infty)$ thus for $0\le x\leq 5$ we have $0\leq x^2\leq 25\leq 30$ – R.N Nov 11 '15 at 22:36

Looks like you are right.

If $x\in\mathbb{Z}$, that implies $x$ is an integer. If $x\in\mathbb{R}$, that implies $x$ is a real number. Know the difference.

Looking at (a), if $x \le 30$, and $x\in\mathbb{Z}$, then $-5 \le x \le 5$, since $6^2 = 36 \ge 30$. Thus, (a) is false.

Note the slight difference in part (b). $x \le 30$, but this time $x\in\mathbb{R}$. Since $x$ can be any real number, $x \ge 5$. In fact, the inequality follows that $-\sqrt{30} \le x \le \sqrt{30}$. Thus, (b) is false.

Looking at (c), if $x \le 5$, then $x^2 \le 25$. However, since $x\in\mathbb{Z}$, $x^2$ could be greater than $30$ (i.e. $x = -6$). Therefore, part (c) is false.

Looking at (d), if $x \le 5$, then $x^2 \le 25$. Therefore, $x \le 30$. This is similar to the question in part (c).

Hope this helped. Comment if you have questions.

• I'm confused as why you concluded (a) is false. Surely $-5\le x \le 5$ implies $x\le 5$. See the Comment by @Takhteh_pareh about this a few seconds before your Answer posted. – hardmath Jun 20 '16 at 17:49