Find the flaw with the following proof that every postage of

Find the flaw with the following proof that every postage of 3 cents or more can be formed using just 3-cent and 4-cent stamps.

Basis step: We can form postage of 3 cents with a single 3-cent stamp and we can form postage of 4 cents using a single 4-cent stamp.

Inductive step: Assume that we can form postage of j cents for all integers j with 3 j k using just 3-cent and 4-cent stamps. We can then form postage of k + 1 cents by replacing one 3-cent stamp with a 4-cent stamp or by replacing two 4-cent stamps with three 3-cent stamps.

Solution

3 cents can be made by using one 3-cent stamp.

Now lets say that a postage of N cents can be made, say by a 3-cent stamps and b 4-cent stamps.

=> , we have
3a + 4b = N.
Then (a-1) 3 - cent stamps and (b+1) 4 - cent stamps will give a postage of
3(a-1) + 4(b+1) = 3a - 3 + 4b + 4 = (3a + 4b) + 1

                                                         = N + 1 cents.
therefore , by mathematical induction,

the above explanation shows that every postage of 3 cents or more can be formed using 3 - cent and 4 - cent stamps.

but the flaw lies in the fact that if we let a = 0

in this case the induction process fails.