# All for One

In 1988, Florida International University mathematician T.I. Ramsamujh offered a proof that all positive integers are equal. “The proof is of course fallacious but the error is so nicely hidden that the task of locating it becomes an interesting exercise.”

Let p(n) be the proposition, ‘If the maximum of two positive integers is n then the integers are equal.’ We will first show that p(n) is true for each positive integer. Observe that p(1) is true, because if the maximum of two positive integers is 1 then both integers must be 1, and so they are equal. Now assume that p(n) is true and let u and v be positive integers with maximum n + 1. Then the maximum of u – 1 and v – 1 is n. Since p(n) is true it follows that u – 1 = v – 1. Thus u = v and so p(n + 1) is true. Hence p(n) implies p(n+ 1) for each positive integer n. By the principle of mathematical induction it now follows that p(n) is true for each positive integer n.

Now let x and y be any two positive integers. Take n to be the maximum of x and y. Since p(n) is true it follows that x = y.

“We have thus shown that any two positive integers are equal. Where is the error?”

(T.I. Ramsamujh, “72.14 A Paradox: (1) All Positive Integers Are Equal,” Mathematical Gazette 72:460 [June 1988], 113.)

(The error is explained here and here.)