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r/learnmath•
unpythonic

[High School Pre-Calc] Does the Pythagorean theorem imply its converse?

I flipped through my niece's Pre-Calculus book today and came across a proof that I didn't think was correct. I thought I had learned the Pythagorean theorem as:

If a right triangle has legs a and b and hypotenuse c then a^2 + b^2 = c^2 .

The proof in this book depends on the Pythagorean theorem saying the converse:

If a triangle has sides a, b and c such that a^2 + b^2 = c^2 then it is a right triangle.

While it is trivial to use the Law of Cosines to show this second statement is true, it isn't clear to me that the Pythagorean theorem says this. I looked at the Wikipedia and Wolfram pages on it, and didn't come away thinking that the Pythagorean theorem was biconditional. What did I miss?

Posted Jan 11, 2013, 6:50 AM UTC

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kempff•

The official version of the Pythagorean Theorem goes like this:

  • For a triangle, IF it is right THEN it has a^2 + b^2 = c^2

Thus the Pythagorean Theorem is a statement about right triangles.

The converse goes:

  • For a triangle, IF it has a^2 + b^2 = c^2 THEN it is right

This is not a statement about right triangles, but a statement about triangles that have a certain relation of sides.

The converse is not the official version of the Pythagorean Theorem, but lots of people lump them together in their heads because it is legitimate to construct the biconditional that arises herefrom:

  • For a triangle, being right and having a^2 + b^2 = c^2 are either simultaneously true or simultaneously false and never obtain separately

EDIT: The proof in the textbook relies on the converse.

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unpythonicOP•

What bothers me is the proof saying "you can conclude by the Pythagorean theorem that the triangle must be right" when they are using the converse of a theorem that is not stated as a biconditional.

Is it legitimate to "conclude by the Pythagorean theorem" something which the Pythagorean theorem doesn't say? It seems like one would need to do a small proof of the converse in the margins along the lines of:

Let a and b be the lengths of two legs of right triangle T. By PT the length of the hypotenuse, c, is sqrt(a^2 + b^2 ). By SSS, we know that any triangle with sides a,b,c will be congruent to T. Therefore all triangles with sides a,b and c are right. QED.

With the converse proven, one would now be free to use it.

I feel like if I had written a proof like the one in the book in one my upper division math courses, I would have been severely penalized for violating so fundamental a tenet of formal logic.

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kempff•

Modern textbooks are notoriously sloppy in this regard. No, you can conclude by the converse of the PT that the triangle is right. The textbook is assuming that the biconditional is the "Pythagorean Theorem", which is historically inaccurate; for nevertheless that's not how Euclid did it (cf. http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI47.html and http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI48.html).

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tunstall•

What bothers me is the proof saying "you can conclude by the Pythagorean theorem that the triangle must be right" when they are using the converse of a theorem that is not stated as a biconditional.

But it is stated as a biconditional on the previous page. (in parantheses, but still, it does say it)

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unpythonicOP•

Interesting. Google books (what I linked) that I was using to flip through the book randomly chooses to show some pages and hide others. When I was looking at page 5 last night, Google had hidden page 4. Today I'm using a browser on a different computer and... there it is. I guess given that they state "the converse is also true" then the proof they provide is satisfactory.

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hobbes727•

Think of it this way for the converse...any triangle with sides a, b, and c can only form one triangle. Therefore if a^2 + b^2 = c^2 the only possible configuration of the sides is a right triangle by the Pythagorean theorem.

It may seem like cyclical logic, but it is a result of a, b, and c forming a unique triangle.

Edit: formatting

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Bobshayd•

More rigorously, if a^2 + b^2 = c^2, then a right triangle with legs length a and b has hypotenuse c, and therefore is congruent to our triangle. Since our triangle is congruent to a right triangle, it is right.

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unpythonicOP•

Think of it this way for the converse...any triangle with sides a, b, and c can only form one triangle.

No argument here.

Therefore if a^2 + b^2 = c^2 the only possible configuration of the sides is a right triangle by the Pythagorean theorem.

Maybe it will help if I use propositional logic:

P: "A right triangle has legs a, b and hypotenuse c"
Q: "a^2 + b^2 = c^2 "

The way I had written the Pythagorean Theorem, which seems to agree with what I have read on Wikipedia and Wolfram, is:

P -> Q

It may seem like cyclical logic, but it is a result of a, b, and c forming a unique triangle.

I don't think it's circular logic, it's affirming the consequent (or my premise statement of the theorem is wrong, which I'm trying to understand).

We could posit that perhaps there exists some a,b,c with this property that forms a unique triangle, but not a right triangle. This doesn't violate the Pythagorean theorem as stated because ~P -> Q is still true.

The error is either my statement of the theorem (it's incomplete or stated too informally) or the proof in the book isn't rigorous.

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[deleted]•

Suppose that a particular right triangle does not satisfy a^2 + b^2 = c^(2), then it follows that the -2abcos(C) term is non-zero, and therefore C is not 90 degrees. This contradicts the original supposition of our triangle being right, and so completes the proof by showing it is indeed bidirectional, that is a^2 + b^2 = c^2 if and only if the triangle is a right triangle.

So really, it can be shown to be a biconditional from just law of cosines stuff, your book just wasn't careful about it.

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j0ma•

Interesting discussion. I always thought of the Pythagorean theorem as a biconditional. However, suppose it were not one, that is:

IF the triangle is right THEN a^2 + b^2 = c^2 Then how exactly would we define that an angle is right, without using the converse of the PT?

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