Is $3\le 11$ a true statement?

Question

$$3(x+1) \le 3x+11$$ simplified: $3 \le 11$.

What is $x$ solution in bracket form?

I think that the expected answer is $(-\infty,\infty)$ a true statement, but it does not feel quite right to me because $3$ is always inferior to $11$ but never equal to it, the sign used in the inequality is $\le $.

What is the right solution in bracket form $(-\infty,\infty)$ or $\emptyset$?

$x\le y$ doesn't exclude $x=y$. For this another relation is used: $y \gvertneqq x$. Or also $x \lneq y$. — Dietrich Burde, Sep 01 '21 at 16:48
"$3\le 11$" means "$3<11$ or $3=11$" which is of course true. — Peter, Sep 01 '21 at 16:49
Since the final inequality holds for every real number, the original inequality does as well, so the solution set is $\mathbb R$ or in interval-form $(-\infty,\infty)$ — Peter, Sep 01 '21 at 16:52
Related: https://math.stackexchange.com/questions/679451/less-than-or-equal-sign — Hans Lundmark, Sep 01 '21 at 18:28

score 1 · Answer 1 · answered Sep 01 '21 at 16:51

The relation $\le$ stands for less than or equal.

That is an inclusive or, thus if any of the two components less than, or equal is true then the compound statement is true.

For example, $3\le5$ is true as well as $5\le5$

The solution for $$3(x+1) ≤ 3x+11$$ is therefore $x\in (-\infty, \infty )$

score 1 · Answer 2 · answered Sep 01 '21 at 16:51

Let's focus on what we mean by a ≤ b. In the sense you used, it is "a equals b and a is smaller than b". But this is a contradiction since there is no a and b so that a is simultaneously equal to and smaller than b. The meaning of "≤" is "smaller or equal". Thus it suffices a to be equal to b OR less than b.

In your example 3 is less than 11 thus it is also true to say 3 ≤ 11. One condition of two is satisfied. 11 ≤ 11 is also correct since again one of two is satisfied.

Is $3\le 11$ a true statement?

2 Answers2