Pseudo Proofs that are intuitively reasonable

What are nice “proofs” of true facts that are not really rigorous but give the right answer and still make sense on some level? Personally, I consider them to be guilty pleasures. Here are examples of what I have in mind:

  1. $s=\sum_{i=0}^\infty \delta^i=1/(1-\delta)$.

    “Proof:” $s=1+\delta+\delta^2+\cdots=1+\delta s$ and hence $s=1/(1-\delta)$.

  2. $(f\circ g)'(x)=f'(g(x))g'(x)$.

    “Proof:” $\displaystyle\lim\limits_{h\to 0}\frac{f(g(x+h))-f(g(x))}{h}=\lim\limits_{h\to 0}\bigg(\frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)}\frac{g(x+h)-g(x)}{h}\bigg).$

  3. $[0,1]$ is uncountable.

    “Proof:” Pick a number from $[0,1]$ randomly. Every number has the same probability. If this probability were positive, there would be finitely many such numbers such that the probability of picking one of them exceeds $1$, which cannot be. So the probability of picking each number is $0$. If $[0,1]$ were countable, the probability of picking any real number would be $0=0+0+0+\cdots$. But by picking from a uniform distribution, I will get a real number with certainty.

It might be helpful to indicate where the lapses in rigor are and why the method works anyways.


Cayley-Hamilton Theorem. Let $A$ be an $n\times n$ matrix, and let $f(t)$ be its characteristic polynomial. Then $f(A)=0$.

Proof.” $f(t) = \det(A-tI)$. Therefore, $f(A) = \det(A-AI) = \det(A-A) = \det(0) = 0$.

Issues. One problem may not be obvious… the equation “$f(A)=0$” is really saying that the matrix we get via the evaluation (by identifying the underlying field with the subring of scalar matrices) is the zero matrix.
However, the “proof” claims to prove that $f(A)$, which is supposed to be a matrix, is equal to the value of a determinant, which is a scalar.

As to why it “gives the right answers”… well, because the theorem is true. I am reminded of what Hendrik Lenstra once said in class after presenting an idea for a proof and explaining why it didn’t quite work:

The problem with incorrect proofs to correct statements is that it is hard to come up with a counterexample.

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