Mathematical Fallacies
An assumption, or series of steps which is theoretically possible but mathematically false is called a Math Fallacy. You'll get around with some examples of fallacies below.
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Contents
Understanding a fallacy
Let's try to prove that \( 2 = 1 \), and state why and where the proof is being wrong.
\(\require{cancel} \text{Assume, } \alpha = \beta \\ \alpha \times \alpha = \beta \times \alpha \text{ (Multiply both side by }\alpha \text{)}\\ \alpha^2 = \alpha \beta \\ \alpha^2 - \beta^2 = \alpha\beta - \beta^2 \text{ (Subtract both side by } \beta^2 \text{)} \\ (\alpha + \beta)(\alpha - \beta) = \beta (\alpha - \beta) \text{ ( Since } a^2 - b^2 = (a+b)(a-b) \text{)} \\ (\alpha + \beta) \cancel{(\alpha - \beta)} = \beta \cancel{(\alpha - \beta)} \\ \alpha + \beta = \beta \\ \alpha + \alpha = \alpha \text{ (Since }\alpha = \beta \text{ )} \\ 2\alpha = \alpha\\ 2\cancel{\alpha} = \cancel{\alpha}\\ 2 = 1\)
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.
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
\(\require{cancel} \text{Let ''e'' be the weight of an Elephant, "m'' be the weight of a mosquito and let their total weight be "a''} \\ \implies e + m = a \\ \implies e = a - m \text{ ... } \fbox{1}\text{ } \\ \text{and } e - a = -m \text{ ... } \fbox{2} \text{ }\\ \text{multiply } \fbox{1} \text{ and } \fbox{2}\\ e (e - m) = -m (a - m) \text{ }\\ \implies e^2 - em = -am + m^2 \text{ }\\ \text{add } (\frac{a}{2})^{2} \text{ on both sides} \\ e^2 - em + (\frac{a}{2})^2 = m^2 - am + (\frac{a}{2})^2 \\ \implies (e - \frac{a}{2})^2 = (m - \frac{a}{2})^2\\ \implies \sqrt[2]{(e - \frac{a}{2})^2} = \sqrt[2]{(m - \frac{a}{2})^2}\\ \implies e - \frac{a}{2} = m - \frac{a}{2} \\ \implies e - \cancel{\frac{a}{2}} = m - \cancel{\frac{a}{2}} \\ \implies e = m \\ \therefore \text{Weight of an elephant is same as weight of the mosquito.} \)
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
Fallacy problems that featured
Here are some "fallacy" contributions from our community.
\[\begin{equation} \begin{split} 1729^2=&1729 \cdot 1729 \quad & \ldots (1) \\ 1729^2-1729^2=&1729 \cdot 1729-1729^2 \quad & \ldots (2) \\ (1729+1729)(1729-1729)=& 1729(1729-1729) \quad & \ldots (3) \\ \require{cancel} (1729+1729)\cdot \cancel{0}=&\require{cancel}1729\cdot \cancel{0} \quad & \ldots (4) \\ 3458=&1729 \end{split} \end{equation} \]
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.
If \(x=y \ne 0 \)
\( \LARGE 1) \) \(x^2 = xy \)
\( \LARGE 2) \) \(x^{2} - y^2 = xy - y^2 \)
\( \LARGE 3) \) \((x+y)(x-y) = y(x-y) \)
\( \LARGE 4) \) \(x + y = y \)
\( \LARGE 5) \) \(x + x = x \), since \(x=y\)
\( \LARGE 6) \) \(x + x = x \)
\( \LARGE 7) \) \(2x = x \)
\( \LARGE 8) \) \(2=1\), since \(x\) is non-zero.
In which "equation" is the first mistake made?
\(1. \quad \frac{0}{0} = \frac{0}{0}\)
\(2. \quad \frac{0}{0} = \frac{100-100}{100-100}\)
\(3. \quad \frac{0}{0} = \frac{(10^2-10^2)}{10(10-10)}\)
\(4. \quad \frac{0}{0} = \frac{(10+10)(10-10)}{10(10-10)}\)
\(5. \quad \frac{0}{0} = \frac{10+10}{10}\)
\(6. \quad \frac{0}{0} = \boxed{2}\)
Consider the following proof.
\(0 = (1-1) + (1-1) + (1-1) + \ldots + (1-1) \rightarrow \fbox{1}\\ \implies 0 = 1 + (-1+1) + (-1+1) + \ldots + (-1+1) \rightarrow \fbox{2}\\ \implies 0 = 1 + 0 + 0 + \ldots + 0 \rightarrow \fbox{3} \\ \implies 0 = 1 \rightarrow \fbox{4}\)
The above is a Mathematical fallacy.
If you know that the proof is wrong, find which step is being wrong.