# Negation of inequalities

So to my understanding, the negation of $$\lt$$ would be $$\ge$$. It's also my understanding that when you negate a false statement, it should be true (and vice versa).

But then I considered this simple situation:

Let a $$\in \{0,1\}$$ and b $$\in \{1,2\}$$

Clearly $$a \lt b$$ is false (i.e. when a = 1 and b = 1)

But the negation $$a \ge b$$ is also false (e.g. when a = 0 and b = 2)

The only correct inequality here $$a \le b$$, but I'm having a hard time trying to put into clear terms why this reasoning breaks down? In particular, in this case why does the negation of a false statement give another false statement?

(For context, I'm currently studying real analysis for the first time, and I've had a lot of proofs which involve inequalities. Through this example, I am starting to realise that maybe proof by contradiction/negation is not the best method when proving an inequality?)

• $a< b$ means nothing. Instead, if you add quantifiers then it has sense. For example, $\exists a\in\{0,1\}\,\exists b\in\{1,2\}\colon a< b$ is true when $a=0$ and $b=1$ (for example). Another statement would be $\forall a\in\{0,1\}\,\forall b\in\{1,2\}\colon a<b$ is false when $a=1$ and $b=1$. But $a<b$ without quantifiers does not mean nothing. Commented Aug 8, 2020 at 8:11

You need quantifiers whenever making statements. $$a by itself is meaningless. Here are two possible statements you could make:

1. $$\forall a\in \{0,1\}$$, $$\forall b\in\{1,2\}$$, $$a
2. $$\exists a\in \{0,1\}$$, $$\exists b\in \{1,2\}$$, $$a

Their negations are

1. $$\exists a\in \{0,1\}$$, $$\exists b\in \{1,2\}$$, $$a\geq b$$
2. $$\forall a\in \{0,1\}$$, $$\forall b\in\{1,2\}$$, $$a\geq b$$

Here, $$(1)$$ is false, $$(2)$$ is true, while $$(3)$$ is true and $$(4)$$ is false.

$$\forall$$ is the logical symbol for "for all" or "for each" or "for every", while $$\exists$$ is the logical symbol for "there exists" (and exists means "there exists atleast one", not "there exists a unique")

You mention real analysis, so you may have seen things like the triangle inequality: "$$|a+b| \leq |a| + |b|$$". In the form currently stated, this is completely meaningless, because I haven't quantified $$a$$ and $$b$$. A meaningful statement (which is also true) is

For every $$a,b\in \Bbb{R}$$, we have $$|a+b| \leq |a| + |b|$$ (with equality if and only if $$a$$ and $$b$$ have the same sign, or one of them is zero)

Its negation below is certainly false:

There exist $$a,b\in \Bbb{R}$$ such that $$|a+b| > |a|+|b|$$.

The moral of the story is that yes it's true the negation of $$<$$ is $$\geq$$, but you also have to make explicit the quantifiers, and negate them.

• Ah that makes a lot of sense! Thank you for the help! Commented Aug 8, 2020 at 8:31
• No problem. You can upvote and accept the answer if it's satisfactory. Commented Aug 8, 2020 at 9:50

You have multiple elements so the way to write possible inequalities is by quantifiers, then negations will work.

Let $$A=\{0,1\},B=\{1,2\}$$. For these sets

$$\exists a\in A ~\forall b\in B~:~ a is a true statement. Its negation needs to be taken by de Morgan's laws.

Its negation is $$\forall a\in A~\exists b\in B~:~a\geq b$$ which is a false statement.

• I see! Thank you! Commented Aug 8, 2020 at 8:31