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My simplest theorem

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Theorem Any natural number that has a divisor greater than itself equals zero.

Proof We observe for any natural n, d, q

n = dq  ∧  d > n

=        { Leibniz }

n = dq  ∧  d > dq

=        { d > 0 }

n = dq  ∧  1 > q

=        { q is natural }

n = dq  ∧  q = 0

⇒        { Leibniz }

n = 0

( End of Proof )

At least twice —EWD1088 & EWD1170— I had used that 0 is the only natural number with infinitely many divisors —e.g. 2k for any k— but I never took the trouble to prove it, and that probably explains why I missed the above.

Austin, 10 February 1996

prof. dr. Edsger W.Dijkstra
Department of Computer Sciences
The University of Texas at Austin
Austin, TX 78712-1188
USA


transcribed by Tristam Brelstaff
modified Mon, 10 Dec 2007.