In this video, I'd like to talk about a common misconception that people have in their first proofs course, and in this example, we are going to do an inequalities example
Question: Show that for all $x,y \in \mathbb{R}$
$$9x^4+4y^2+x^2\geq 12x^2y$$I'm going to present a proof of this fact and the proof will go as follows.
For all $x,y \in \mathbb{R}$, we have:
$$\begin{align*} 9x^4+4y^2+x^2 &\geq 12x^2y\\\\ 9x^4+4y^2-12x^2y+x^2&\geq 0\\\\ 9x^4-12x^2y+4y^2+x^2&\geq 0\\\\ (3x^2-2y)^2+x^2&\geq 0 \end{align*}$$which is true. $\quad\blacksquare$
So here's the proof that I'm presenting to you. What I'd like you to do now is take a minute and just ask yourself is this proof correct or is this proof incorrect.
Hopefully you've done that, and you've critiqued this proof. I mean hopefully you've come to the conclusion this proof is not correct, and the fundamental flaw with this proof is that it's written backwards. Mathematics is read top to bottom, so the way we read this is we start with line $1$, and we say that line $1$ implies that line $2$ is true, which implies that line $3$ is true which implies that line $4$ is true.
However line $1$ is a line that we don't know if it's true or false, that's what the question is asking us to prove. So proving a bunch of implications without knowing that the first term is true is not logically valid as an argument.
There are lots of ways to correct this proof, but I will show you two different ways right now. By the way before I do, again a good exercise is to try to correct this proof for yourself first, and then continue watching the video.
Here's a correct solution to this problem.
For all $x,y \in \mathbb{R}$,
$$(3x^2-2y)^2+x^2 \geq 0$$Since the squares of real numbers is always non-negative. Further this implies that
$$9x^4-12x^2y+4y^2+x^2\geq 0$$and hence
$$9x^4+4y^2+x^2 \geq 12x^2y \quad \blacksquare$$So that works. For another proof, what you can do is you can start with the inequality but instead of having an implication here, which is implicit in the original proof that I gave you, we can make it a bidirectional implication. In other words, an “if and only if” statement.
For all $x,y \in \mathbb{R}$,
$$\begin{align*} 9x^4+4y^2+x^2 \geq 12x^2y &\Leftrightarrow 9x^4-12x^2y+4y^2+x^2\geq 0\\\\ &\Leftrightarrow (3x^2-2y)^2+x^2\geq 0 \end{align*}$$which is true.$\quad\blacksquare$
Why is that valid? Again, we can move a term to the other side and just switch the sign, and we can bring the term back over, and once again just switch the sign, that's okay. So these two things imply each other.
Make sure though that both directions work. So here we're moving terms around, that's okay, you have to be careful though when you're doing things like multiplication and division. One direction might be easy, the other direction might not be possible. For example, we could multiply by $0$ but we can't divide by $0$.
Once you have $9x^4-12x^2y+4y^2+x^2\geq 0$, that is true if and only if $(3x^2-2y)^2+x^2\geq 0$ is true, and again how do we do this? Well we took these three terms, $9x^4-12x^2y+4y^2$, and we factored them, or we took this term, $(3x^2-2y)^2$ and expanded it. Both of those operations are reversible, so that's okay in this situation as well.
The last statement is true, all of these are “if and only if” statements, so therefore this initial statement is true if and only if the second statement is true if and only if the last statement is true.
And so you're done, this is a perfectly valid proof. So again, the key idea here is that we've gone to an “if and only if” statement instead of just an implication. Okay, so hopefully I gave you a little bit of insight into this problem. Thank you very much for your attention.