Each of these proofs result in a false statement. Identify the mistakes.
Let
Multiply by |
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Subtract |
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Factor | |
Divide by |
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Substitute |
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Divide by |
What went wrong?
You can’t divide through by
Start with a true equation | |
Multiply by |
|
Add |
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Reduce 35 and 45 to factors | |
Factorise | |
Take the square root of both sides | |
Subtract |
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:
This is obviously true.
, the calculus way
We know from the power rule that:
However, what if we re-write
Hence
What’s going on here?
We can’t differentiate a sum whose number of terms is dependant on
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
Since the sum is a function of
, integration by parts
Let us evaluate the indefinite integral:
Let:
Thus by integration by parts:
What happened?
This problem illustrates an improper application of integration by parts. When
using the formula, a constant of integration
In reality the last line should read:
Which is trivially true.
Too many solutions
A quadratic has either 0, 1, or 2 unique solutions. Look at this equation:
Without loss of generality we can assume
How can this equation have 3 unique solutions?
The equation is not actually quadratic in
Proof
The original equation:
Multiply by
Expand:
Collecting
Thus all the
Unequal equations
Take the following equation:
Combining the equations:
However, if we substitute into the original equation, we get:
What happened?
The original quadratic has roots at:
However, when we combine the rearranged equations, we form a cubic equation. This adds a solutions, so now:
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