Assume you don't know the result of $(-1)\cdot 3$. You might try calculating $$ \begin{align*} 3\cdot 3=9\\ 2\cdot 3=6\\1\cdot 3=3\\0\cdot 3 = 0 \end{align*} $$ thus you may argue $$(-1)\cdot 3=-3$$ just by recognition of the pattern (first factor decreases by 1, result decreases by 3).

Such argumentations may work, especially when you define something new (like multiplication on negative numbers). However, things may go wrong as cases of $0^0$ show where you could define it as $0$ according to $0^n$ or as $1$ according to the $n^0$-sequence. I would also say that $\int 1/x=ln(x)$ does not fit the pattern of $\int x^z=\frac{x^{z+1}}{z+1}$ for $z \neq -1$. (You might want to add a constant to be exact.)

Question Do you have more of such examples?

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    $\begingroup$ $\int \frac{1}{x} = \log(x) $ does fit the pattern. Specifically, $\displaystyle \int_{t=1}^{t=x} t^z = \frac{x^{z+1} - 1}{z+1}$. The limit of this ratio as $z \to -1^{+}$ is in fact $\log(x)$. $\endgroup$ Commented Jun 16, 2014 at 18:24
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    $\begingroup$ You might find some examples at math.stackexchange.com/questions/111440/… (I suggested ~vonbrand's answer on a different thread that was closed as a duplicate of the aforementioned: math.stackexchange.com/a/819019/37122). $\endgroup$ Commented Jun 16, 2014 at 21:51
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    $\begingroup$ @StevenGubkin I did not mean the integral is wrong in some way, but somewhat unexpected since for any integer other than -1, you get a monomial. The pattern I referred to is like "primitives of monomials are monomials". $\endgroup$
    – Anschewski
    Commented Jun 25, 2014 at 13:58

2 Answers 2


Given $ax + b = 0$ with $a \neq 0$, I can find a formula for $x$. In particular, we have $x = -b/a$.

Similarly, for $ax^2 + bx + c = 0$ with $a \neq 0$, I can also find a formula for $x$. In particular, the quadratic formula gives all possible values for $x$ (there can be up to $2$).

For the cubic case, one can again come up with an analogous formula, which gives all possible values for $x$ (there can be up to $3$). Probably you see where this is going...

For the quartic case, we have a formula (i.e., using the four operations and radicals) to identify the different possible values of $x$ (here, indeed, there can be up to $4$).

By analogy, one would really expect there to be a "quintic formula." But there's not (in a strong sense, i.e., not just because neither you nor I can think of one), and the reasoning by analogy goes wrong.

(Meanwhile, the idea of "possible values" needs to be made precise, since you might end up allowing for complex values of $x$; this notion does generalize nicely, i.e., a polynomial with degree $n$ can have up to $n$ distinct complex roots.)

Of course, the "reasoning by analogy gone wrong" described above is frequently mentioned early on in a class on, say, Galois Theory to motivate its study.

(And speaking of classical problems: Similar comments could be made about trisecting an angle...)

Below is an example that is a bit simpler to explain in full.

Assume you do not know if a two digit number is prime.

Here is a primality test:

First, memorize the twelve by twelve times table.

Second, check if the given number is even (look at units digit), a multiple of three (check if the digital sum is a multiple of three), or a multiple of five (look at units digit).

Third, if the number does not appear in the times table and has no obvious factors of 2, 3, or 5, then it must be prime.

Now, this approach to finding primes does not work so well for larger numbers.

Moreover, it misses a two digit number: 91, the product of 7 and 13.

(Quite a number of students will misidentify 91 as prime; try presenting it in a list of numbers to a student, and ask him or her to circle all the primes given some reasonable time constraint.)

Of course, for numbers through 100 one needs to check prime factors up to its square root, i.e., 2, 3, 5, and 7. But the tricks for 2, 3, and 5 are very simple, and those of us who memorized twelve by twelve tables get the multiples of 7 up through 84 (hence the missed 91, whereas 98 is even).

Still, I think there is a nice example in here somewhere, and I think it is related to the sense that multiplication tables contain "all the products" in a stronger sense than they really do. For example, the ten by ten table is missing all primes from 11 on. As for composites, it is missing plenty, starting at 22; it just so happens that the primality test described above is effective for all two-digit numbers besides 91.

  • $\begingroup$ Not sure I agree with the "give a formula for $x$" example. Those formulas have virtually no visual similarities with each other, so really, the only thing they share is existing. I don't know if a student who saw the linear, quadratic, cubic, and quartic formulas put beside each other would believe that anyone could manage to find a fifth one. In a sense, the natural analogy would lead to the correct answer (and in an even vaguer sense, for the correct reason: lack of symmetry). $\endgroup$
    – Ryan Reich
    Commented Jun 18, 2014 at 15:15
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    $\begingroup$ John H. Conway likes to say that 91 is the smallest number that looks prime but isn't. $\endgroup$ Commented Jun 18, 2014 at 15:47
  • $\begingroup$ @RyanReich The similarity is exactly what you identify: existence. I think the unsolvability of the quintic is very surprising, given that there are formulae for polynomials of degree 1, 2, 3, and 4. My sense is most students would guess that a formula exists, but that it is very complicated; even the idea of proving non-existence can be tough to grapple with after years of using the quadratic equation. $\endgroup$ Commented Jun 19, 2014 at 6:46
  • $\begingroup$ @MikeShulman Accepting $91$ as prime leads to questioning unique facotrization using the example $1001= 11*91=7*143$. :-) $\endgroup$
    – Anschewski
    Commented Jun 25, 2014 at 14:11
  • $\begingroup$ @Anschewski but clearly 143 is not prime, since it's divisible by 11. $\endgroup$ Commented Jun 26, 2014 at 5:37

The polynomial $n^2 + n + 41$ gives only prime values for $0 \le n \le 40$, while for 40 it is $41 \cdot 41$.

  • $\begingroup$ More like a failure of applying the inductive reasoning? $\endgroup$ Commented Mar 7, 2023 at 4:05

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