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.
12141655110112151316718
12141655110112151316718

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.
1
1

However, it turns out that there is an n for which P(n) is true. n = 4700063497 is the first one!
0
0
