Understanding Spivak’s answer Chapter 1, 16(b) and the role of contradiction in a proof

algebra-precalculuscalculusinequality

Using the fact that ${x}^{2} + 2xy + {y}^{2} = (x + y)^{2} \ge 0$ , show that $4{x}^{2} + 6xy + 4{y}^{2} > 0$ unless $x$ and $y$ are both $0$

This has been asked and answered here, but not in the spirit that Spivak has asked.

Spivak's answer:

The first equation implies that $$4{x}^{2} + 8xy + 4{y}^{2} \ge 0$$ Suppose we also had $$4{x}^{2} + 6xy + 4{y}^{2} \le 0$$ Subtracting the second from the first would give us $2xy \ge 0$. If neither $x$ nor $y$ is $0$, this means we must have $2xy>0$; But this implies that $4{x}^{2} + 6xy + 4{y}^{2} > 0$, a contradiction.
Moreover, it is clear that if one of $x$ and $y$ is $0$, but not the other then we also have $4{x}^{2} + 6xy + 4{y}^{2} > 0$

My analysis…

By subtracting the second from the first equation, we are able to isolate $xy$, and we show that $2xy \ge 0$. Now, to expand a little on Spivak, if

neither $x$ nor $y$ is $0$, then either $x,y > 0 $, or $x,y < 0$ (in order to obtain $2xy > 0$)

Applying that to $4{x}^{2} + 6xy + 4{y}^{2}$ gives us the desired answer that $4{x}^{2} + 6xy + 4{y}^{2} > 0$.
And, to finish off, if either $x$ or $y$ is $0$, we are left with $4y^{2} > 0$ or $4x^{2} > 0$

…and a question

Assuming the analysis is correct, I would like to understand more fully the issue of contradiction

Suppose we also had $4{x}^{2} + 6xy + 4{y}^{2} \le 0$

this means we must have $2xy>0$; But this implies that $4{x}^{2} + 6xy + 4{y}^{2} > 0$, a contradiction.

Does this contradiction imply the opposite is true? ie because ($4{x}^{2} + 6xy + 4{y}^{2} \le 0$) is false, then ($4{x}^{2} + 6xy + 4{y}^{2} > 0$) must be true? (Even though we go from $\le 0$ to $> 0$)

Best Answer

Suppose we also had $4{x}^{2} + 6xy + 4{y}^{2} \le 0$

this means we must have $2xy>0$; But this implies that $4{x}^{2} + 6xy + 4{y}^{2} > 0$, a contradiction.

Does this contradiction imply the opposite is true? ie because ($4{x}^{2} + 6xy + 4{y}^{2} \le 0$) is false, then $(4{x}^{2} + 6xy + 4{y}^{2} > 0)$ must be true? (Even though we go from $\le 0$ to $> 0$)

Since our entire argument has been valid, the contradiction (absurdity or logical falsity) that has been derived must have arisen from the premise/assumption; in other words, our assumption must be false; in other words, its negation must be true.

I think calling the negation "opposite" is potentially ambiguous, and possibly why you ask the next question:

$(4{x}^{2} + 6xy + 4{y}^{2} > 0)$ must be true? (Even though we go from $\le 0$ to $> 0$)

The negation of $a>b$ is $a\leq b,$ while its "opposite" might be reasonably construed as $a<b.$

P.S. In the given exercise, the proof's premise/assumption is being contradicted, leading us to conclude that it is false. So, in fact, this proof would have been more succinctly cast as a contrapositive proof. The contradiction that is derived in a proof by contradiction does not in general involve the deliberately false assumption. More here.

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