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What's the flaw in the following “proof” by strong induction that every postage of 3 cents or more can be formed using only 3-cent and 4-cent stamps?

P(k): postage of k cents can be formed using only 3-cent and 4-cent stamps. Basis step: P(3): one 3-cent stamp. P(4): one 4-cent stamp.

Inductive step:

Assume P(j) is true for all positive integers 3 ≤ j ≤ k (Inductive Hypothesis).

To obtain postage for k + 1 cents we can consider the postage for k cents (by Inductive Hypothesis) and either replace one 3-cent stamp with a 4-cent stamp OR by replacing two 4-cent stamps with three 3-cent stamps.

Thus P(k+1) is true.

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    $\begingroup$ A good way to find a flaw in an induction proof is to look at the first case where it fails and then see where the induction step goes wrong in that case. $\endgroup$ – Eric Wofsey Nov 5 '19 at 3:56
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You cannot do "replace one 3-cent stamp with a 4-cent stamp" when $k$ is $4$.

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  • $\begingroup$ but if we have one 3 cent stamp then, this should do it for k = 4 right? $\endgroup$ – Xander Nov 5 '19 at 7:13
  • $\begingroup$ @Xander Yes, but actually no. The key is if we have one 3-cent stamp. But you cannot use one 3-cent stamp plus some 4-cent stamp to form 4-cent. The only way is to use one 4-cent stamp. $\endgroup$ – edm Nov 5 '19 at 7:18
  • $\begingroup$ but we can't use one 4 cent stamp? cuz if thats the case then, we cant get k = 3 either? cuz you cant have one 3 cent + 4 cent to get 3? $\endgroup$ – Xander Nov 6 '19 at 19:44
  • $\begingroup$ @Xander 3 cent is formed with exactly one 3-cent stamp and no 4-cent stamp, 4 cent is formed with exactly one 4-cent stamp and no 3-cent stamp. $\endgroup$ – edm Nov 7 '19 at 1:28
  • $\begingroup$ yeah so when k = 3, we one one 3 cent stamp, but when k = 4, you replace that one 3 cent stamp with one 4 cent stamp, so it should work ? $\endgroup$ – Xander Nov 8 '19 at 21:20
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Consider the case $k=4$! The only possibility in this case is one 4-cent stamp!

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