0

I decided to try proof by induction without any help = ) so if someone could check it out, pretty sure it's unfinished or, well i'm not sure. Also, if possible, could you take a logical guess for how many marks i'd get /10 for what I've done atm :

Consider the function five: N > N define recursively as follows: 1) base case: five(0) = 10 2) recursive case: five(x) = five(x-1)+5 for any x>0

prove using induction on the natural numbers that the following equality is true for all natural umbers n e(subset) N: five(n)=(n+2)*5

My working out

Check if it holds for n1 five(1)=five(1-0)+5 = 10 + 5 = 15

Assume K=N

five(k) = five(k-1)+5 proving this holds

five(k+1)=five(k+1-1)+5

At this point I got a bit lost, but soldiered on :D

five(1) = five(1)+5 five(2) = five(2)+5

that's as far as I got, any help much appreciated

user118964
  • 11

2 Answers2

0

when you check for 1, and then assume that your general equation (n+2)*5 holds for n and check for n+1, this is how you should do it.

out of the definition of five, we know the following follows : five(n + 1) = five (n) + 5. We know our general equation (n+2)*5 = five(n) holds for n and below, therefore we can substitute five(n) giving

five(n+1) = (n+2)5 + 5 = 5 (n + 3) = 5 * ( (n+1) + 2).

So, our equation checks out.

You were on the right path before you got lost, remember that you assume your general formulla holds for "n and smaller". When proving with induction, always try to express f(n+1) in function of f(n), substitute your general formula for f(n) and see if that formula holds for f(n+1).

camel
  • 170
0

Hint: Your assumption says that $$five(n)=5(n+2)$$ Thus, for the inductive step you have $$\begin{align*}five(n+1)&=five(n)+5\\ &=5(n+2)+5 \end{align*}$$ Can you take it the last bit?

E.O.
  • 6,942
  • 5(1+2)+5 = five(1-0)+5=10+5=15 – user118964 Jan 03 '14 at 23:50
  • I'm not really sure what i'm supposed to do tbh from where you left off – user118964 Jan 03 '14 at 23:51
  • @user118964 your hypothesis is that $five(n)=5(n+2)$. You proved it for the base case, now you need to show that assuming the statement for $n$ implies that it holds for $n+1$. That basically means you need to get that last expression to look like $5((n+1)+2)$ and you are done. – E.O. Jan 04 '14 at 00:00