What are some false proofs for true or false statements where the error in the proof is not obvious?

Question

I'm looking for examples of subtle errors in reasoning in a mathematical proof. An example of a 'false' proof would be

Let $a=b>0$. Then $a^2 - b^2 = ab - b^2$. Factoring, we have $(a-b)(a+b) = b(a-b)$, which after cancellation yields $b = a = 2b$ and thus $1=2$.

However the divide-by-zero trick is trite once you've seen it enough. Are there any other 'near miss' proofs (can be of anything) whose error in reasoning is harder to spot?

Edit: This question was put on hold for being too opinion based, so I would like to try to clarify. I'm looking for false proofs of true or false statements with subtle logic errors, which for me would consist of everything except basic algebra errors (dividing by zero, $\sqrt{ab} = \sqrt{a}\sqrt{b}$ for negative $a,b$, etc.) and basic logic errors ($(p\Rightarrow q) \implies (q \Rightarrow p)$).

Answer 1

This is a logical proof.

Theorem: Either a statement or its converse must be true.

Proof: Consider the proposition $$(P\to Q)\vee (Q\to P)$$ and check that it's a tautology, say using truth tables. The point is that if $P$ is false, then $P\to Q$ is true, and similarly if $Q$ us false, then the second implication holds. Finally, if both are true, then both implications hold. $\Box$

For example, you could take $P$ to be that $x$ is a prime number, and $Q$ to be that $x$ is odd. Then we are claiming either that $x$ is prime implies it is odd or that $x$ is odd implies that it is prime. (Neither is true!) What's going on?

The point is that the theorem is technically true without quantifiers. However in the example I gave, there were quantifiers and it really should have been written $$\forall x[P(x)\to Q(x)]\vee \forall x[Q(x)\to P(x)],$$ which is not the same as $$\forall x[(P(x)\to Q(x))\vee (Q(x)\to P(x))].$$ This was one of the more interesting examples from this book.

Answer 2

A conceptual error commonly made-

$({\sqrt -1})$$=-1^\frac12$$=-1^\frac24$$=(-1^2)^\frac14$$=1^\frac14$$=1 $

Answer 3

$\cos^2x=1-\sin^2x$

$\cos x = (1-\sin^2x)^{\frac{1}{2}}$

$1+\cos x = 1+(1-\sin^2x)^{\frac{1}{2}}.$

$When,x=180^0$

$1-1 = 1+(1-0)^{\frac{1}{2}}$ $as$$,$$[$$cos 180=-1$ $and$ $sin 180=0$$]$

$or,$$0=2$

Answer 4

There is this false induction type of thing, where the step does not work in all cases.

Theorem. If there is one white horse, all horses are white.

Proof. Induction on the number $n$ of horses existing. If there is only one horse $(n=1)$, and as there is one white horse, it is white. Now suppose, there are $n>1$ horses, $h_1, \ldots, h_n$ with - say - $h_1$ white. Looking only at $h_1, \ldots, h_{n-1}$, by induction, all these are white. But know, under $h_2, \ldots, h_n$ we have white horses, and again, by induction, all are white.

Answer 5

There is a famous old "proof" that all triangles are equilateral. I first came across it in Harold Stark's An Introduction to Number Theory (pp. 223-224). Carlo Sèquin gives a nice, five-minute youtube demonstration of it.

Answer 6

The following deduction is not correct:

$i^2 = i*i = {\sqrt{-1}} * {\sqrt{-1}} = {\sqrt{(-1) * (-1)}} = {\sqrt{1}} = 1$