Can you give me a good example of a mathematical theorem that appears to be true but is actually false?

Question

My students always ask me why proofs are needed in math. I am looking for a theorem that works for many cases and seems to be always true but actually has a counterexample proving it false. This will demonstrate why a formal proof is needed in mathematics.

Answer 1

Here's a good statement:
"Every continuous curve in the plane has zero area".
It looks like it should be true. But it is FALSE. In fact,there is a continuous curve that fills a whole square! This curve 'intersects itself' infinitely often, so we might then conjecture
"Every arc (a one-to-one curve) has zero area".
But this is FALSE also!
Finding an example of an arc with strictly positive area is not easy, but it can be done (the usual examples require the use of Cantor sets).

Depending on the level, you might be able to say:
"Every continuous function is differentiable except at finitely many points".
This is also FALSE. There are continuous functions that are differentiable NOWHERE.
Or you might try:
"Every differentiable function is monotone on some interval".
Again, this is FALSE.
Or another:
"A function with derivative zero almost everywhere is constant".
The 'Devil's staircase' is a counterexample.

Answer 2

First, one should be careful with the language as we don't allow false statements to be mathematical theorems. So strictly speaking, the answer to your question is "No", we can't find a mathematical theorem that appears to be true but is actually false.

Of course, what you really mean to ask, I presume, is for a mathematical statement that appears true but is actually false. The Jay H's answer is an excellent one. A related problem (though not as nice) is that of Fermat primes. In case you are unfamiliar, Fermat claimed that numbers of the form 2^(2^r)+1 are always prime. One can check for small r, namely 0,1,2,3,4, that his assertion is correct. However, for r>= 5, there are no known Fermat primes. While this may not seem particularly relavent to anything, it turns out to have a use in geometry, namely an n-gon is constructible with straightedge and compass if and only if the odd primes dividing n are Fermat primes. This, for instance, is how you can prove that a 17-gon is constructible (a result first found by Gauss).

I'm not sure at what level you are teaching, but another idea (which doesn't answer your question directly) is to consider matrices. It is an easy excercise to come up with two 2x2 matrices that do not commute (i.e., AB does not equal BA). This is counterintuitive since the numbers with which we traditionally work do commute. This might motivate the need for proving the commutative laws of addition and multiplication for whatever number system you are using (e.g., integers, rational numbers, real numbers, complex numbers, etc.,)

Hope this helps.

Answer 3

How about the internal angles of a triangle add up to 180 degrees. However, if you start at the North pole, travel South 100 miles in a straight line, turn 90 degrees to the right travel 100 miles West in a straight line and turn 90 degrees to the right again and then travel 100 miles North in a straight line you'll end up at the North pole at a different angle from where you started. So, how come the angles all added up come to more than 180 degrees when you have walked around a triangle?

You can obviously do it so that the total internal angle add up to any values more or less but it can be done where every angle is 90 degrees which is easy enough to do if you travel a quarter of the way around the latitude circle at which you end up after the first leg.

We teach Euclidian geometry so exclusively at school that it's often forgotten how much it depends on the world being flat!

Answer 4

I guess superbotz has given you a good idea, but here is where his proof falls through:

x = y = 1
x^2 = xy
x^2 - y^2 = xy - y^2
(x+y)(x-y) = y(x-y)

now we need to note that x-y is 0, this now gives us

(x+y)0 = y(0)
0 = 0, which is true for every x and y. So this proof is flawed because it overlooks some basic priciples. Just for interest sakes, the prof that 1 not= 0 is:
Let x not= 0 be given. Assume 1 = 0
implies x = x.1 = x.0 = 0
but x not= 0
therefor 1 not= 0, by contradiction

Answer 5

How about the proof that 1 = 0?

Take the following identity:
x = y = 1
Now, multiply all sides by x:
x² = xy
Subtract y² from each side:
x² - y² = xy - y²
Factor each side:
(x+y)(x-y) = y(x-y)

(At this point is what I call a "conditional equality". I'm not sure if this is a real conditional equality or not. My definition is an equality that is only true because of certain terms. If those terms were removed, then the equality would be false.)

Now, divide by (x-y) (See what I mean? It divides by 0.)
x + y = y
Subtract y from both sidesz
x = 0
Therefore, 1 = 0.

Answer 6

How about the so-called theorem that f(n) = n^2 − n + 41 seems to yield primes for n = 0, 1, 2, ... but stops working when n = 41?

EDITED TO ADD: Superobotz' example is just the opposite of what you're looking for -- it's a flawed proof.

Answer 7

What is 0^0 (zero raised to the zeroeth power)?

A naive case can be made for 0^0 = 1, since
. lim x^0 = 1
x → 0

Call 0^0 = 1 your "theorem".

However, if you just stop there you miss out! Another naive case can be made for 0^0 = 0, since
. lim 0^y = 0
y → 0

So which is it? In point of fact, neither 1 nor 0 can be "proven" to be the right answer, because it CAN be proven that
. . . lim x^y . . . does not exist.
(x,y)→(0,0)

(by using the analysis we just did in fact)

Answer 8

Er,maybe Fermats last theorem.Or 1=0 And i wonder how does those ancient mathematician started their great ideas on mathematic theorem and proved them?How does Euler and Gauss thinks?Was it their imagination or logic?

Answer 9

In laptop language && returns genuine purely whilst the two are appropriate, the situation returns genuine if any of them is stable. Its certainty table is a million acts as genuine & 0 acts as fake For and nil && 0 = 0 0 && a million = 0 a million && 0 = 0 a million && a million = a million For OR 0 0 = 0 00 0 = 0 a million