Bogus proof of relation between two non negative real numbers

Question

$a$ and $b$ are two non negative real numbers and I need to find a relation between them. I claim that $a\ge b$, and propose this proof,

Bogus proof.

Suppose that $a\ge b$,

$$a\ge b\tag*{so}$$ $$a-b\ge 0\tag*{so}$$ $$\left( a-b\right)^2\ge 0\tag*{which we know is true.}$$

This proves the claim.$\tag*{$\blacksquare$}$

What's wrong with the above proof?

Answer 1

The main error is that you start with what you want to proof and derives something true, and then conclude that what you started with must be true. But $A \implies B$ where $B$ is true does not mean that $A$ must be true. For that you need a reverse implication: $A \Longleftarrow B$.

So, the assumption $x \geq 0 \implies x^2 \geq 0$ which indeed is true, but $x \geq 0 \Longleftarrow x^2 \geq 0$ is not a true implication.

Answer 2

$P\implies Q$ does not mean that $Q\implies P$.

In this case specifically, $a\ge b \implies (a-b)^2\ge0$ does not imply $(a-b)^2\ge 0 \implies a\ge b$.

This shows that the argument is invalid.

In this case the conclusion is false since $(a-b)^2 \ge 0$ does not imply $a-b\ge 0$.

Answer 3

The general problem of this proof is the following: If $p$ and $q$ are two statements, then $p\Rightarrow q$ is especially true, if $p$ is false - no matter if $q$ is true or false. Therefore, the short answer to your question is: you are disregarding the order of proof (start with a true statement and then use logical deduction to arrive at a true statement).

If you look for the precise mistake in your calculations, you need to find the place, where your "so" can not be replaced by an "if and only if". Can you now pinpoint, what your mistake is?

Answer 4

The significant problem is that not every step of your proof is reversible. Indeed, taking $$a\geq b \implies a-b\geq 0$$is a reversible step (that is to say, $a-b\geq 0 \implies a\geq b $) and therefore totally legitimate in a proof.

However, the squaring step is irreversible. Notice that $$A\geq 0\implies A^2\geq 0$$ but $$A^2\geq 0\implies |A|\geq 0$$ the slight difference makes this an illegitimate step.