How is a general case equivalent to a special case and how showing a special case demonstrates a general case, in the proof of pythagoras theorem?

Question

"The general theorem expressed by $\lambda a^2 = \lambda b^2 + \lambda c^2 $ is equivalent not only to the special case $a^2 = b^2 + c^2 $ but to any other special case. Therefore, if any such special case should turn out to be obvious, the general case would be demonstrated." (Ref. Chapter 2, section 5 of Mathematics and plausible reasoning vol 1. by George Polya )

Questions

1. What is meant by the statement above that a general case is equivalent to a special case?
2. How can I demonstrate the general case by choosing a obvious special case? Wouldn't it be wrong in the sense that it is similar to demonstrating that $5^2 = 3^2 + 4^2$ and thus concluding that $a^2 = b^2 + c^2$ in general?

Answer 1

In context, the special case $a^2 = b^2 + c^2$ is referring to Pythagoras' theorem with squares on the sides of the right triangle. The general case $\lambda a^2 = \lambda b^2 + \lambda c^2$ is referring to arbitrary but similar shapes draw on the sides of the right triangle. The equivalence comes from reasoning about the areas of similar shapes.

To answer your questions:

1. Here 'equivalent' means 'logically equivalent'. The special case with squares implies the general case with arbitrary shapes (because of how similarity works), and vice-versa.

2. Without any extra information, you can't demonstrate a general case from a special case. But in this context, Polya is talking about a general case (arbitrary shape) that has been shown to be logically equivalent to any special case (particular shape). So showing that a special case is true (pythagoras true for a particular shape) demonstrates the general case (pythagoras true for an arbitrary shape) which in turn demonstrates the original special case (pythagoras is true for squares).

Answer 2

1. It means that there is not a (simple) demonstration that $a^2+b^2=c^2$ implies $\lambda a^2=\lambda b^2+\lambda c^2$, namely just multiply both sides by $\lambda$ and expand on the right. (Of course there is also the derivation of the special case from the general case by just plugging in $\lambda =1$ and simplifying).

2. There are general cases and there are general cases. You talk about $5^2=3^2+4^2$ as a special case of $a^2=b^2+c^2$, whereas the quote talks about $a^2=b^2+c^2$ as special case of $\lambda a^2=\lambda b^2+\lambda c^2$. The claim that "any special case true implies general case true" works only if (as above) the general case is equivalent to the special case.

Answer 3

(Picture needed, but we will use words instead)

Consider $\triangle ABC$, right-angled at $C$. Drop a perpendicular from $C$ to $AB$, meeting $AB$ at $P$.

Then triangles $CBP$, $ACP$, and $ABC$ are similar, and have hypotenuses $a$, $b$, and $c$ respectively.

It is obvious that in area, $\triangle CBP+\triangle ACP=\triangle ABC$.

There is a constant $\mu$ such that $\triangle ABC$ has area $\mu c^2$. By similarity, the area of $\triangle CBP$ is $\mu a^2$, and the area of $\triangle ACP$ is equal to $\mu b^2$. Thus $\mu a^2+\mu b^2=\mu c^2$, and we obtain the Pythagorean Theorem.

The general theorem is that if similar figures are drawn on the legs and hypotenuse of a right triangle, then the sum of the areas of the figures on the legs is equal to the area of the figure on the hypotenuse. The obvious special case is the triangles $CBP$, $ACP$, and $ABC$, where the fact that the sum of the first two is equal to the third is clear. And from this obvious special case we can easily derive the general case, and in particular the case of classical interest, where the figures are squares.