An assumption or series of steps which is seemingly correct but contains a flawed argument is called a mathematical fallacy. This page contains some examples of mathematical fallacies below.
Let's try to prove that 2=1, and state why and where the proof is wrong.
Assume, α=βα×α=β×α (Multiply both side by α)α2=αβα2−β2=αβ−β2 (Subtract both side by β2)(α+β)(α−β)=β(α−β) ( Since a2−b2=(a+b)(a−b))(α+β)(α−β)=β(α−β)α+β=βα+α=α (Since α=β )2α=α2α=α2=1
In the above proof, you'll be amused how can this happen. We knew 2 is not equal to 1, so somewhere the proof is wrong. If you look at the 5th step, (α−β) is being cancelled by dividing (α−β) on both the sides. Here is where the proof is wrong. Let's see why.
α=β=>α−β=0 .... (1)(α−β)(α−β)=00=undetermined.
So, in the 5th step we cannot divide it by (a−b), as 00 cannot be determined.
How to find it, where does it go wrong?
You can easily find you have a fallacy in your statement if you idenitfy the following results in your math-script.
Your proof is being theoretically correct, and no mistakes are found
Your proof ended with some equals, which are universally unequal. For example, 2 = 1, a = b, where a > b etc..,.
Other types of fallacies and Examples
Before we see other types of fallacies, let's take a look at interesting example.
Weight of an elephant = Weight of a mosquito
Let ”e” be the weight of an Elephant, "m” be the weight of a mosquito and let their total weight be "a”⟹e+m=a⟹e=a−m ... 1and e−a=−m ... 2multiply 1 and 2e(e−a)=−m(a−m)⟹e2−ea=−am+m2add (2a)2 on both sidese2−ea+(2a)2=m2−am+(2a)2⟹(e−2a)2=(m−2a)2⟹2(e−2a)2=2(m−2a)2⟹e−2a=m−2a⟹e−2a=m−2a⟹e=m∴Weight of an elephant is same as weight of the mosquito.
The fallacy happened in the last-third step in which two square roots are being cancelled.
a2=m⟹a=±2m,So while having two square roots, a2=b2; we must make it clear that ±a=±b So not only a = b, but -a = b, a = -b too.
So in the above step, we can only assume that ∣e−2a∣=∣m−2a∣
Here is another type of fallacy.
Consider any variable "x" such that x> 0. So for ’x’ we have, x=x⟹x=x times1+1+.....+1⟹x2=x timesx+x+.....+x⟹dxdx2=dxdx timesx+x+.....+x differentiate on both sides.⟹2x=x times1+1+.....+1⟹2x=x Since, 1. x times 1 is x. 2. 1 + 1 + .... + 1 is already equal to x in second step.⟹2=1 !!
The above shows my attempt to prove that 3458=1729. In which of these steps did I make a flaw in my logic?
Study the following steps. All of the statements below are correct except for one (or more). Which one is the first fallacious step? Indicate your answer with the number of the statement you think is errorneous.