Proof of $\forall$ statement given interval notation.

Question

I'm working on my discrete math homework, and I'm blanking on how to write a proof for a logical statement involving the quantifier $\forall$ The statement I need to prove is as follows: $\forall x\in \Bbb R,$ if $x \in [1,2],$ then $(3x-1) \in [2,5]$ We worked on some of those in class today, but as always, I didn't seem to take very strong notes, and can't remember what to do now. Thanks in advance for all the help.

Answer 1

Expanding upon ConMan's answer, here is a full proof:

Consider $x\in [1, 2]$. Then, $1\leq x\leq 2\implies 3\leq 3x\leq 6\implies 2\leq 3x-1\leq 5.$

Answer 2

The general way to prove a "for all" statement is:

1. Take an arbitrary element of the condition.

2. Prove that is satisfies the conclusion.

So in this case, a proof would look something like this:

Let $x \in [1, 2]$. Then $x \geq 1$ and $x \leq 2$.

...

Therefore, $3x - 1 \in [2, 5]$.

Answer 3

$x\in [1,2] \iff 1 \le x \le 2$

$1\le x \le 2$ : Hypothesis

$3 \le 3x \le 6$ : Multiplication law with positive factor

$2 \le 3x-1\le 5 \iff (3x-1) \in [2,5]$: additive law for inequalities.

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Adding a constant to either side of an inequality preserves the inequality.

Multiplying both sides by a positive number also preserves the inequality. Multiplying by a negative number reverses the inequality.

So for example:

$1\le x \le 2$: Hypothesis

$-6 \le -3x \le -3$: Negative Factor Multiplication for inequalities

$-5 \le -3x+1 \le -2$: Same old Additive law for inequalities.

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