Each of these proofs result in a false statement. Identify the mistakes.
\(2 = 1\)
Let \(a\) and \(b\) be non-zero numbers.
| \(a = b\) | |
| \(a^2 = a b\) | Multiply by \(a\) |
| \(a^2 - b^2 = ab - b^2\) | Subtract \(b^2\) |
| \((a-b) (a + b) = b (a - b)\) | Factor |
| \(a + b = b\) | Divide by \(a-b\) |
| \(2 b = b\) | Substitute \(a = b\) |
| \(2 = 1\) | Divide by \(b\) |
What went wrong?
You can’t divide through by \(a-b\). Since \(a = b\), this is dividing by 0.
\(4 = 5\)
| \(20 = 20\) | Start with a true equation |
| \(36 - 16 = 45 - 25\) | |
| \(16 - 36 = 25 - 45\) | Multiply by \(-1\) |
| \(16 - 36 + \frac{81}{4}= 25 - 45 + \frac{81}{4}\) | Add \(\frac{81}{4}\) to both sides |
| \(4^2- 36 + \left( \frac{9}{2} \right)^2 = 5^2 - 45 + \left( \frac{9}{2} \right)^2\) | |
| \(4^2 - \left( 2 \times4 \times \frac{9}{2} \right) + \left( \frac{9}{2} \right)^2 = 5^2 - \left( 2 \times 5 \times \frac{9}{2} \right) + \left( \frac{9}{2} \right)^2\) | Reduce 35 and 45 to factors |
| \(\left( 4 - \frac{9}{2} \right)^2 = \left( 5 - \frac{9}{2}\right )^2\) | Factorise |
| \(4 - \frac{9}{2} = 5 - \frac{9}{2}\) | Take the square root of both sides |
| \(4 = 5\) | Subtract \(\frac{9}{2}\) |
What happened?
Square roots have 2 solutions in general. When taking the square root in the proof, on one side we took the negative square root and on the other side we took the positive square root.
If on the other hand we took the positive square root on both sides we would have:
\[\begin{align} - \left(4 - \frac{9}{2}\right) & = 5 - \frac{9}{2} \\ \frac{1}{2} & = \frac{1}{2} \end{align}\]This is obviously true.
\(2 = 1\), the calculus way
We know from the power rule that: \(\frac{d}{dx} x^2 = 2 x\)
However, what if we re-write \(x^2\) as the sum of \(x\)s:
\[\begin{align} \frac{d}{dx} x^2 & = \frac{d}{dx} (x + x + x + \ldots + x) \\ & = \frac{d}{dx} x + \frac{d}{dx} x + \frac{d}{dx} x + \ldots + \frac{d}{dx} x \\ & = 1 + 1 + 1 + \ldots + 1 \\ & = x \end{align}\]Hence \(2x = x\) for any \(x\), therefore \(2 = 1\).
What’s going on here?
We can’t differentiate a sum whose number of terms is dependant on \(x\) by simply taking the sum of the derivatives. To make the fallacy here more explicit, write out the sum formally:
\[f(x) = x^2 = \sum_{i=0}^x x\]There are a few fallacies here. Firstly this equation is not meaningful for non-integers. Functions are only differentiable on a continuous space such as the real numbers.
For the second fallacy, note that in the proof, we took the derivative with respect to each term in the sum, but not with respect to the number of terms. The number of terms is a function of \(x\) and has to be taken into account.
Since the sum is a function of \(x\), we must use the chain rule when evaluating this. However this approach does not get us very far. Let the function that the sum denotes be called \(g(x)\). Unsurprisingly we find that \(g(x) = x^2\), leaving us with the same problem.
\(0 = 1\), integration by parts
Let us evaluate the indefinite integral: \(\int \frac{1}{x} dx\)
Let:
\[\begin{align} u &= \frac{1}{x} &dv &= dx &\\ du &= - \frac{1}{x^2} dx &v &= x& \end{align}\]Thus by integration by parts:
\[\begin{align} \int \frac{1}{x} dx & = \frac{x}{x} - \int \frac{1}{x^2}x dx \\ \int \frac{1}{x} dx & = 1 + \int \frac{1}{x} dx \\ 0 & = 1 \end{align}\]What happened?
This problem illustrates an improper application of integration by parts. When using the formula, a constant of integration \(C\) must be added. Up the second last line the equations are correct. However, the last line does not follow, as you cannot cancel \(\int \frac{1}{x}dx\), because they are not necessarily equal. There are an infinite number of antiderivatives of \(\frac{1}{x}\), which all differ by a constant factor.
In reality the last line should read:
\[0 = 1 + C\]Which is trivially true.
Too many solutions
A quadratic has either 0, 1, or 2 unique solutions. Look at this equation:
\[\frac{(x-a)(x-b)}{(c-a)(c-b)} + \frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-c)(x-a)}{(b-a)(b-c)} = 1\]Without loss of generality we can assume \(a < b < c\). Now note that \(x = a\), \(x = b\) and \(x = c\) are all unique solutions.
How can this equation have 3 unique solutions?
The equation is not actually quadratic in \(x\). It doesn’t depend on \(x\) at all, and hence has an infinite number of solutions for \(x\).
Proof
The original equation:
\[\frac{(x-a)(x-b)}{(c-a)(c-b)} + \frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-c)(x-a)}{(b-a)(b-c)} = 1\]Multiply by \((a-b)(b-c)(a-c)\):
\[(x-a)(x-b)(a-b) + (x-b)(x-c)(b-c) + (x-a)(x-c)(a-c) = (a-b)(b-c)(a-c)\]Expand:
\[(x^2 - (a + b) x + ab)(a-b) + (x^2 - (b + c) x + bc)(b-c) - (x^2 - (a + c) x + ac)(a-c) = (a-b)(b-c)(a-c)\]Collecting \(x\) terms:
\[x^2 (a - b + b - c - a + c) + x (a^2 - b^2 + b^2 - c^2 -a^2 + c^2) + ab(a-b) + bc(b-c) - ac(a-c) = (a-b)(b-c)(a-c)\]Thus all the \(x\) terms cancel out, leaving an equation that is always true.
Unequal equations
Take the following equation:
\[x^2 + x = -1\]\(x = 0\) is not a solution so therefore we can divide by \(x\) and rearrange:
\[\begin{align} x + 1 & = - \frac{1}{x} \notag \\ \frac{1}{x} + x & = -1 \end{align}\]Combining the equations:
\[\begin{align} x^2 + x & = \frac{1}{x} + x \\ x^2 & = \frac{1}{x} \\ x^3 & = 1 \\ x & = 1 \end{align}\]However, if we substitute into the original equation, we get:
\[1^2 + 1 = -1\]What happened?
The original quadratic has roots at:
\[x = \frac{-1 \pm \sqrt{3}}{2}\]However, when we combine the rearranged equations, we form a cubic equation. This adds a solutions, so now:
\[x = \frac{-1 \pm \sqrt{3}}{2}, 1\]And indeed, these are the three cube roots of one.
In general setting two equations equal to each other can generate more solutions, that don’t satisfy the original equations independently. This is because you are removing constraints, in this case, the constraint that both equations equaled \(-1\).