my homework do it

Prove by induction that each positive integer n ≥ 8 can be represented in the form
n = 3a + 5b, where a and b are non-negative integers.

no

3a + 5b = 3a + 3b + 2b
= 3(a+b) + 2b, with a and b an element of N
= 3c + 2(c-a), with a and c an element of N
= 3c + 2c - 2a
= 5c - 2a = 1 for c = 1 and a = 2 -> you can make steps of 1 from n = 3*1 + 5*1 = 8

QED

>>3
At step 3, you forgot "and (c - a) an element of N", in other words, c >= a. This is why the conclusion in step 5 is wrong - you can't express n=1 with the given constraints.

Real answer:
n = {8, 9, 10} can be solved by hand. "n >= 11 is expressible" follows from n-3 being expressible.

-> you can make steps of 1 from n = 3*1 + 5*1 = 8

Wat. I don't get the final line.

>>5
Probably because it's wrong.

>>4
so n-3 = 3(a-1) + 5b

how is it clear a-1 is never negative for n>=11?

>>7
wait, fuck it, i don't even know if i'm on the right track with this.

>>7
For n=8, 11, 14, 17, etc:
First we prove that it holds for n=8. Simple: n=3*1+5*1.
Now, let's assume 8+3x (x being a non-negative integer) can be represented as 3a+5b. Then 8+3(x+1) = 8+3x+3 can be represented as 3a+5b+3 = 3(a+1)+5b:

8+3(x+1)
= (taking 3 outside the parentheses)
8+3x+3
= (by assumption (formal term: "induction hypothesis"))
(3a+5b)+3
= (taking 3 inside the parentheses)
3(a+1)+5b

That's it, we're done (for n=8, 11, 14, 17, etc). We've shown that the condition we're interested in holds for n=8 (n=8+0x), and we've shown that IF it holds for n=8+3x, THEN it also holds for n=8+3(x+1). This is enough to conclude that it holds for n=8+3x, for all (non-negative integer) x.

Constructing a similar proof for the other two cases is left as an exercise to the reader.

Thanks. I'm not even OP btw, I just read his problem and it got me interested.

fail where n=9

>>11
a=3, b=0
non-negative integers