I don't know the full process of induction, as it's one of the harder questions on my upcoming test papers I thought i'd attempt to get a basic understanding; in order to get a 1/2 marks out of the question. However, since doing it I believe with a bit of help I could get full marks. Here's my attempt, I know it's not great but.. if someone could point me in the right direction of where i'm going wrong, i'd be most grateful.
I know I have posted the same question above, but the difference being today I've actually attempted it using a method I found on google so would like someone to take a look
$\text{b)}$ Consider the function $\rm take:\mathbf N\to\mathbf N$ defined recursively as follows:
$\qquad1)$ Base case: $\,\,\rm take(0)=100$
$\qquad2)$ Recursive case: $\,\,\,\rm for \,\,any\,\,x\gt0,\,\,\, take (x)=take(x-1)-2$Prove using induction on the natural numbers that the following equality is true for all natural numbers $\rm n\in\mathbf N$: $$\rm take(n)=100-(2*n)$$
take (0) = 100 for any x>0, take(x)=take(x-1)-2
take(n)=100-(2*n)
1) 1>0 take(1)= take(1-1)-2
show true for n=1
take(0)=100 take(1)=take(1-1)-2 =take(0)-2 =100-2 =98
Assume it's true for n=k
take(k)=take(k-k)-2 = true
therefore
- take(k+1)=take(k+1-2)-2 = true
- take(1+1)=take(k)-2
- take(2)=take(1)-2
- take(2) = take(k)-2 due to our assumption
- take(k)=take(k-k)-2
- take(2)=take(k-k)-2
at the end I've got no idea where I was going with it tbh...
You make the inductive assumption that $\;T(n)=100-2n\;$ , with $\;T=$ the "take" function, and then you must prove it for $\;n+1\;$ :
$$T(n+1):\stackrel{\text{By def.}}= T(n)-2\stackrel{\text{Ind. Hypothesis}}=\left[100-(2n)\right]-2=100-2(n+1)$$
and the rigthmost expression is exactly what it must be when using $\;n+1\;$ instead of $\;n\;$ in the definition of $\;T\;$ , so we're done!