A=B implies that 1=1, therefore…

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I’ve ranted in the past about the fallacy of trying to prove an identity by starting with the equation itself, then manipulating both sides of the equation until you arrive at a valid identity.

While grading some homework this term (involving proofs of trig identities), I found the need to raise the subject again in class.  But my stock example, proving that -3 = 3 by squaring both sides, seemed too transparent.  I wanted something where the fallacy was solely due to proving that False implies True.

I ended up with the following example, which I like a lot, but which I’m certain has been rediscovered by others over the ages.  Still, it’s a good illustration of why we can’t prove identities in this way.

Claim: \sin x = \cos x, for all x.

Proof: Assume that \sin x = \cos x.  If we square both sides, this implies that \sin^2 x = \cos^2 x.  Furthermore, since equality is reflexive symmetric, it follows that \cos^2 x = \sin^2 x.

Finally, adding these two equations gives us \sin^2 x + \cos^2 x = \cos^2 x + \sin^2 x, which reduces to the equation 1=1.  QED.

The careful reader will note that squaring both sides is irrelevant, as is the Pythagorean identity for sine and cosine.  In essence we have a general proof that A = B for any expressions A and B:  If A = B, then by reflexivity symmetry we know that B=A, and thus A + B = B + A.  But since addition is commutative, this reduces to the identity A+B = A+B.

I prefer the slight obfuscation of the \sin^2 x + \cos^2 x proof over the distilled simplicity of “reflexive symmetric plus commutative”.

Your mileage may vary.

12 Responses to “A=B implies that 1=1, therefore…”

  1. Denise Says:

    I like your proof!

    It seems to me that the only legitimate kind of proof that can start with “Assume (fill in the blank)” is a proof by contradiction. In other words, you have to start by assuming the opposite of what you want to prove. Otherwise, all you have is a nonsense loop. I can “prove” anything is true, if I’m allowed to start by assuming that it’s true.

    This is why we teach students to work with only one side of the equation (as mentioned in your prior rant), so that they do not inadvertently indulge in circular reasoning. Or rather, the “work with only one side” rule is easier for students to understand and follow than “each step must involve an if-and-only-if inference.”

  2. Dave Richeson Says:

    TwoPi, I’m glad to hear that this bothers you too! I wrote a blog post about this same issue not long ago. I see this type of “proof” all the time, and it really bothers me. Quite a few readers commented on the post. It was an interesting discussion.

  3. Brent Says:

    Nice!

    By the way, I think you mean “symmetric” instead of “reflexive”. Reflexivity is the property that A = A for all A.

  4. TwoPi Says:

    Dave: Wow, what a great post and great discussion! That Lakatos gets mentioned (relevantly) in the comments is an indicator of their breadth and depth. Clearly this is a raw nerve for many.

    Brent: Thank you, yes. Not sure what my brain was doing. Perhaps that’s why my students rolled their eyes when I said that in class? (I’d assumed it was their ingrained dread of equivalence relations, which I may have to blog about in the near future…)

  5. Ξ Says:

    I had an interesting experience and insight when I presented this “proof” to by Abstract Algebra class. I’d just seen a bunch of this kind of reasoning in their homework, so I started class by writing 3 “proofs” on the board: this one, the proof that 5=-5 (set them equal, square both sides), and a proof from the homework.

    These are students I’ve taught in several classes, so their familiar with my personal bugaboos (this being one), and they immediately knew what I was trying to point out (You can’t prove things this way). But because they were seniors and grad students, we were able to have a more detailed conversation about this kind of proof. One thing that came up is that putting a question mark above the equal sign doesn’t solve the problem, and another was that yes, there are proofs that use the result, but in every case (contradiction or contrapositive) you’re assuming the result ISN’T true, not that it is.

    But the most interesting thing to me was the realization that a source of confusion for some students was that in my counterexamples — I’ve used the 5=-5 faithfully over the years — the initial statement is something they already know to be false, whereas on homework they are typically trying to prove something that they know [from the “Prove that…” wording of the problem] to be true. In other words, if you know that what you’re trying to prove is true, is it OK to prove it like this?

    I hadn’t thought of this particular aspect from a student perspective before — to me, I know that I don’t “know” the truth of a statement until I’ve proved it, and sometimes I honestly don’t know the truth at all and that’s what I’m trying to figure out. But my students are rarely uncertain of the end result, and while they have seen “Prove or find a counterexample” I don’t think they’ve ever personally been in a situation where they thought something was true, “proved” it by saying it’s true, and only later found out it wasn’t. My faithful counterexamples from that perspective seem contrived.

    There’s some discussion about this in your blog post, Dave. At this point, I’m still mulling over how to help with the this distinction between a false statement and a false proof. [Some of which just takes seeing more proofs — I don’t think I could have had this conversation with students who were much earlier in their mathematical coursework because they would have a harder time articulating their own assumptions.]

  6. LSquared Says:

    Hmm…What if you just wrote your own homework problems by taking a bunch of the same things to prove from the book, added a few false ones, and changed the instructions to: prove or disprove. Then they don’t know it’s true.

  7. Ξ Says:

    LSquared…that’s a good point. This comes up often enough that that would be quite useful to create, too.

  8. jd2718 Says:

    Hmm. But if you did not use a later step that depended on the hypothesis… And if each step were reversible… Because that’s how I write all my cute inductions (and then reverse the steps before showing them to anyone).

    Jonathan

  9. Brent Yorgey Says:

    It doesn’t matter whether you start with something you already “know” to be true or not, this sort of “proof” is still not valid (unless all the intermediate steps are “if and only if” steps). A valid proof has to convince someone who *doesn’t* already “know” that the statement is true!

    I’m a big fan of the “prove or disprove” type of question; it’s much closer to what math is really about! Even better (but appropriate only at a certain level) are questions of the form “state and prove something interesting about X”.

  10. jd2718 Says:

    Just like back in school! I think that my reversible steps are equivalent to iffs.

    And I never got in trouble for writing proofs backwards and presenting them forwards, until I confessed that that’s what I’d been doing… After that, I just stopped sharing my “process.”

    Jonathan

  11. Brent Yorgey Says:

    “Backwards” reasoning (sometimes called “backwards chaining”) is a perfectly legitimate proof technique; you just have to be careful about which way the implications go! Informally, it corresponds to saying things like “OK, I have to prove X. Now in order for X to be true, it would suffice to show Y (that is, Y implies X), so now I have to show Y. Well, Y is true if Z and Q are both true…” and so on.

  12. Carnival of Mathematics #62 | John D. Cook Says:

    […] you know that sine and cosine are equal for all x? Heather (Xi) submitted a pseudo-proof in A=B implies that 1=1, therefore? by her colleague TwoPi at 360. (If there is ever a 360th Carnival of Mathematics, Heather should […]

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