As someone that finds mathematical proof fascinating, I thought I would write a post on common fallacies in maths.

A mathematical fallacy is a typical error committed during a proof or an argument. Fallacies do not necessarily mean that the conclusion is incorrect, but rather that the argument is incorrect, meaning that the conclusions reached can be surprising. To demonstrate this I will go through classic examples of fallacies.

**1 = 2 (using algebra)**

Let *a*=*b*.

Then a^{2} = ab

a^{2} + a^{2 }= ab + a^{2}

2a^{2} = a^{2 }+ ab

2a^{2} – 2ab = a^{2 }+ ab – 2ab

2a^{2} – 2ab = a^{2} – ab

2(a^{2} – ab) = (a^{2} – ab)

Cancelling the (a^{2} – ab) from both sides gives 1 = 2.

*So, what’s the problem? *From the first step we know that a = b, therefore a^{2} – ab = 0. When we ‘cancel’ the a^{2} – ab, we are essentially dividing both sides by a^{2} – ab. However, we cannot divide by 0, so the proof contains a fallacy!

**Infinite Series**

Suppose `S` = 1 − 1 + 1 − 1 + 1 − 1 + 1 − …

`S`= (1 − 1) + (1 − 1) + (1 − 1) + …

`S`= 0 + 0 + 0 + 0 + …

`S`= 0

and

`S`= 1 + (1 − 1) + (1 − 1) + (1 − 1) + …

`S`= 1 + 0 + 0 + 0 + 0 + …

`S`= 1

**A Ladder will Fall Infinitely Fast when Pulled**

Let *x* denote the horizontal distance from the bottom of the ladder to the wall* *and let *y* denote the height of the top of the ladder from the ground, at time *t**.*

Since the ladder, the ground and the wall from a right hand triangle:. Therefore, it follows that .

Differentiating this gives us

where x’ and y’ represent the derivatives of x and y with respect to *t**.*

Since the bottom of the ladder is being pulled at a constant speed v, x’ = v.

As x approaches L, the numerator approaches -Lv (non zero) whilst the denominator approaches zero, so y’ approaches infinity as x approaches L.

*Where’s the fallacy? *The fallacy is when we say that , as there is an assumption that the top of the ladder remains resting against the wall, however once the ladder has reached a sufficiently small angle to the horizontal, the top will pull away from the wall too, so there is no longer the relationship .

**All People in a Group are the Same Age?**

**Statement S(n):** In any group of

*n*people, everyone in that group has the same age. We will attempt to prove this by induction

- In any group that contains one person, everyone in that group has the same age, therefore statement S(1) is true.
- Assume that when n = k, S(n) is true.
- Let G be an arbitrary group containing k+1 people. We need to show that everyone in G has the same age. To show this, we just need to show that, if
*P*and*Q*are any members of*G*, then they have the same age. - Consider everybody in
*G**except**P*. These people form a group of*k*people, so they must all have the same age, since we assumed that, in any group of*k*people, everyone has the same age (in step 2). - Consider everybody in
*G**except**Q*. Again, they form a group of*k*people, so they must all have the same age. - Let
*R*be someone else in*G*other than*P*or*Q*. Since*Q*and*R*each belong to the group in step 4, they are the same age and since*P*and*R*each belong to the group considered in step 5, they are the same age. - Since
*Q*and*R*are the same age and*P*and*R*are the same age, it follows that*P*and*Q*are the same age and so everyone in*G*has the same age. - So S(n) is true for n = k+1 when n = k is true. As it’s true for n= 1, it must be true for all
*n*.

The fallacy in this proof is in step 6. If G is a group of k+1 people, as long as *k *> 1 (*k*+1 > 2) and in this case there does exist a third person *R* in *G *so the rest of the proof will work. Therefore, this proof shows that *S*(*k*) implies *S*(*k*+1), *as long as k>1,* which in other words means that it shows that

*if*

*S*(2) is true, then so is

*S*(3), and if

*S*(3) is true, so is

*S*(4), etc. However, we have not shown that

*S*(2) is true. The basis step of the induction showed that

*S*(1) was true, but our induction step does not show that

*S*(1) implies

*S*(2).

Hope you enjoyed this post. Friday’s post will be about famous paradoxes! M x

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