John Travis
Mississippi College
November 2022
An interesting idea is to check a conditional proposition P(n) for a bunch of values of n and if true (or false) for all of those then you might presume that P(n) has the same truth value for all natural numbers n. Of course, this is not a proof but just a hunch.
To see why this may be a bad approach (using your gut), consider the following statement;
$P(n): \exists n \in N \ni 2^n \equiv_n 3 $
The cell below allows you to check this statement for several modestly sized values of n....that is, values of n for which you could do this by hand. To check, we will instead compute $2^n - 3$ and see if the remainder is 0. If so, then P(n) above is true. If the remainder is not 0, then P(n) is false.
1 2 1 4 1 6 5 5 1 10 1 12 1 5 13 16 7 18 1 2 1 4 1 6 5 5 1 10 1 12 1 5 13 16 7 18 |
Notice that none of these equals zero and thus for none of these values of n does the original P(n) become true.
Below, take a try by plugging in any value of n you want and see if you can find one where P(n) is true by seeing where the output is zero.
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However, it turns out that there is an n for which P(n) is true. n = 4700063497 is the first one!
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