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ipynb/Beal.ipynb
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@ -33,44 +33,34 @@
"\n", "\n",
"# Online Beal Counterexample Checker\n", "# Online Beal Counterexample Checker\n",
"\n", "\n",
"Got a counterexample? Check it with my **[Online Beal Counterexample Checker](http://norvig.com/bealcheck.html)**.\n", "Got a counterexample? Check it with my [Online Beal Counterexample Checker](http://norvig.com/bealcheck.html).\n",
"\n", "\n",
"# How to Not Win A Million Dollars\n", "# How to Not Win A Million Dollars\n",
"\n", "\n",
"\n", "\n",
"* A proof must show that there are **no** examples that satisfy the conditions. A common error is to show how a certain pattern generates an infinite number of $(A, x, B, y, C, z)$ examples, and that the conjecture holds for this entire infinite collection. But that's not good enough, unless you can also prove that the conjecture holds for every other possible pattern.\n", "* A proof must show that there are **no** examples that satisfy the conditions. A common error is to show how a certain pattern generates an infinite number of $(A, x, B, y, C, z)$ examples, and that the conjecture holds for this entire infinite collection. But that's not good enough, unless you can also prove that the conjecture holds for every other possible pattern.\n",
"\n", "\n",
"<p>\n",
"\n", "\n",
"* It is valid to use proof by contradiction: assume the conjecture is true, and show that that leads to a contradiction. It is not valid to use proof by circular reasoning: assume the conjecture is true, put in some irrelevant steps, and show that it follows that the conjecture is true.\n", "* It is valid to use proof by contradiction: assume the conjecture is true, and show that that leads to a contradiction. It is not valid to use proof by circular reasoning: assume the conjecture is true, put in some irrelevant steps, and show that it follows that the conjecture is true.\n",
"\n", "\n",
"\n", "\n",
"<p>\n",
"\n",
"* A valid counterexample needs to satisfy all four conditions&mdash;don't leave one out.\n", "* A valid counterexample needs to satisfy all four conditions&mdash;don't leave one out.\n",
"\n", "\n",
"\n", "\n",
"<p>\n",
"\n",
"* One correspondent claimed that $27^4 + 162 ^ 3 = 9 ^ 7$ was a solution, because the first three conditions hold, and the common factor is 9, which isn't a prime. But of course, if $A, B, C$ have 9 as a common factor, then they also have 3, and 3 is prime. \"No common prime factor\" means the same thing as \"no common factor greater than 1.\"\n", "* One correspondent claimed that $27^4 + 162 ^ 3 = 9 ^ 7$ was a solution, because the first three conditions hold, and the common factor is 9, which isn't a prime. But of course, if $A, B, C$ have 9 as a common factor, then they also have 3, and 3 is prime. \"No common prime factor\" means the same thing as \"no common factor greater than 1.\"\n",
"\n", "\n",
"<p>\n",
"\n", "\n",
"* Another claimed that $2^3+2^3=2^4$ was a counterexample, because all the bases are 2, which is prime, and prime numbers have no prime factors. But that's not true; a prime number has itself as a factor.\n", "* Another claimed that $2^3+2^3=2^4$ was a counterexample, because all the bases are 2, which is prime, and prime numbers have no prime factors. But that's not true; a prime number has itself as a factor.\n",
"\n", "\n",
"<p>\n",
"\n", "\n",
"* A creative person offered $1359072^4 - 940896^4 = 137998080^3$, which fails both because $3^3 2^5 11^2$ is a common factor, and because it has a subtraction rather than an addition (although, as Julius Jacobsen pointed out, it could be rewritten as $137998080^3 + 940896^4 = 1359072^4$).\n", "* A creative person offered $1359072^4 - 940896^4 = 137998080^3$, which fails both because $3^3 2^5 11^2$ is a common factor, and because it has a subtraction rather than an addition (although, as Julius Jacobsen pointed out, it could be rewritten as $137998080^3 + 940896^4 = 1359072^4$).\n",
"\n", "\n",
"<p>\n",
"\n", "\n",
"* Mustafa Pehlivan came up with an example involving 76-million-digit numbers, which took some work to prove wrong (using modulo arithmetic). \n", "* Mustafa Pehlivan came up with an example involving 76-million-digit numbers, which took some work to prove wrong (using modulo arithmetic). \n",
"\n", "\n",
"<p>\n",
"\n", "\n",
"* Another Beal fan started by saying \"Let $C = 43$ and $z = 3$. Since $43 = 21 + 22$, we have $43^3 = (21^3 + 22^3).$\" But of course $(a + b)^3 \\ne (a^3 + b^3)$. This fallacy is called [the freshman's dream](https://en.wikipedia.org/wiki/Freshman%27s_dream) (although I remember having different dreams as a freshman).\n", "* Another Beal fan started by saying \"Let $C = 43$ and $z = 3$. Since $43 = 21 + 22$, we have $43^3 = (21^3 + 22^3)$.\" But of course $(a + b)^3 \\ne (a^3 + b^3)$. This fallacy is called [the freshman's dream](https://en.wikipedia.org/wiki/Freshman%27s_dream) (although I remember having different dreams as a freshman).\n",
"\n", "\n",
"<p>\n",
"\n", "\n",
"* Multiple people proposed counterexamples similar to this one:" "* Multiple people proposed counterexamples similar to this one:"
] ]