Mathematical Fallacies

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.

$\text{Assume, } \alpha = \beta \\ \alpha \times \alpha = \beta \times \alpha \text{ (Multiply both side by }\alpha \text{)}\\ \alpha^2 = \alpha \beta \\ \alpha^2 - \beta^2 = \alpha\beta - \beta^2 \text{ (Subtract both side by } \beta^2 \text{)} \\ (\alpha + \beta)(\alpha - \beta) = \beta (\alpha - \beta) \text{ ( Since } a^2 - b^2 = (a+b)(a-b) \text{)} \\ (\alpha + \beta) \cancel{(\alpha - \beta)} = \beta \cancel{(\alpha - \beta)} \\ \alpha + \beta = \beta \\ \alpha + \alpha = \alpha \text{ (Since }\alpha = \beta \text{ )} \\ 2\alpha = \alpha\\ 2\cancel{\alpha} = \cancel{\alpha}\\ 2 = 1$

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 $5^{th}$ step, $(\alpha - \beta)$ is being cancelled by dividing $(\alpha-\beta)$ on both the sides. Here is where the proof is wrong. Let's see why.

$\alpha = \beta \\ => \alpha - \beta = 0 \text{ .... (1)} \\ \frac{(\alpha - \beta)}{(\alpha - \beta)} = \frac{0}{0} = undetermined.$

So, in the $5^{th}$ step we cannot divide it by $(a-b)$, as $\frac{0}{0}$ cannot be determined.

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..,.

Before we see other types of fallacies, let's take a look at interesting example.

Weight of an elephant = Weight of a mosquito

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$\text{Let ''e'' be the weight of an Elephant, "m'' be the weight of a mosquito and let their total weight be "a''} \\ \implies e + m = a \\ \implies e = a - m \text{ ... } \fbox{1}\text{ } \\ \text{and } e - a = -m \text{ ... } \fbox{2} \text{ }\\ \text{multiply } \fbox{1} \text{ and } \fbox{2}\\ e (e - a) = -m (a - m) \text{ }\\ \implies e^2 - ea = -am + m^2 \text{ }\\ \text{add } (\frac{a}{2})^{2} \text{ on both sides} \\ e^2 - ea + (\frac{a}{2})^2 = m^2 - am + (\frac{a}{2})^2 \\ \implies (e - \frac{a}{2})^2 = (m - \frac{a}{2})^2\\ \implies \sqrt[2]{(e - \frac{a}{2})^2} = \sqrt[2]{(m - \frac{a}{2})^2}\\ \implies e - \frac{a}{2} = m - \frac{a}{2} \\ \implies e - \cancel{\frac{a}{2}} = m - \cancel{\frac{a}{2}} \\ \implies e = m \\ \therefore \text{Weight of an elephant is same as weight of the mosquito.}$

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The fallacy happened in the last-third step in which two square roots are being cancelled.

$a^2 = m \\ \implies a = \pm \sqrt[2]{m}, \\ \text{So while having two square roots, } a^2 = b^2 \text{; we must make it clear that } \pm a = \pm b \text{ So not only a = b, but -a = b, a = -b too.}$

So in the above step, we can only assume that $|e - \frac{a}{2}| = |m - \frac{a}{2}|$

Here is another type of fallacy.

$\text{Consider any variable "x" such that x> 0. So for 'x' we have, } \\ x = x\\ \implies x = \underbrace{1 + 1 + ..... + 1}_\text{x times} \\ \implies x^2 = \underbrace{x + x + ..... + x}_\text{x times} \\ \implies \frac{d}{dx} x^2 = \frac{d}{dx} \underbrace{x + x + ..... + x}_\text{x times} \text{ differentiate on both sides.} \\ \implies 2x = \underbrace{1 + 1 + ..... + 1}_\text{x times} \\ \implies 2x = x \text{ Since, 1. x times 1 is x. 2. 1 + 1 + .... + 1 is already equal to x in second step.} \\ \implies 2 = 1 \text{ !!}$

Now let's prove that $1 = -1$

$-1 = -1 \\ \implies \frac{1}{-1} = \frac{-1}{1} \\ \implies \sqrt[2]{\frac{1}{-1}} = \sqrt[2]{\frac{-1}{1}} \\ \implies \frac{\sqrt[2]{1}}{\sqrt[2]{-1}} = \frac{\sqrt[2]{-1}}{\sqrt[2]{1}} \\ \implies \sqrt[2]{1}\sqrt[2]{1} = \sqrt[2]{-1}\sqrt[2]{-1} \\ \implies 1 = -1 \text{ !! :-)}$

To get to know where this went wrong, grab a look at Complex Numbers

Here are some "fallacy" contributions from our community.