Question from my book.
Prove that $n^2 7n +1$ for all $n \geq 8 $.
The result is true when $n=8$ because $8^2 = 64$ and $7x8 +1 = 57$. Suppose it is true when $n$ is any number $k \geq 8$, that is $k^2 7k+1$. Then
$$ (k+1)^2 = k^2 +2k +1 (7k+1) +2k+1 = 7(k+1)+1+(2k-6)$$
Since $k \geq 8$, $2k-6$ is a natural number and the last expression above is greater than $7(k+1)+1$. The induction step is verified, and so the result is true for all $n \geq 8$.

Could someone explain this for me? How is this a valid proof? It is supposed to be a about induction.

Question

Question from my book.
Prove that $n^2 > 7n +1$ for all $n \geq 8 $.
The result is true when $n=8$ because $8^2 = 64$ and $7x8 +1 = 57$. Suppose it is true when $n$ is any number $k \geq 8$, that is $k^2 > 7k+1$. Then 
$$ (k+1)^2 = k^2 +2k +1 > (7k+1) +2k+1 = 7(k+1)+1+(2k-6)$$
Since $k \geq 8$, $2k-6$ is a natural number and the last expression above is greater than $7(k+1)+1$. The induction step is verified, and so the result is true for all $n \geq 8$.

Could someone explain this for me? How is this a valid proof? It is supposed to be a about induction.

mr.math · Answer

Read this http://cms.dt.uh.edu/faculty/delavinae/F05/2305/Induction.PDF

slaaibak · Answer

You first have to proof for n=8, the base case.  Now if this is true, you need to assume it's true for n.

Then you have to prove it for n+1

When you prove it for n+1, you prove it for 9, then 9 becomes n, so you prove it for 9+1=10

This process goes on until infinity.  Therefore it proves for all n>=8

mr.math · Answer

I think his question is why the principle of mathematical induction is true. In the link I gave above you can find the proof of the principle theorem.

anonymous · Answer

I don't argue with induction. just with this problem, how does this line make sense "Since k≥8, 2k−6 is a natural number and the last expression above is greater than 7(k+1)+1. "

anonymous · Answer

@Mr.Math reading it now

zarkon · Answer

2*k-6>0 for k>=8

mr.math · Answer

Oh so you're just talking about this problem, sorry!

zarkon · Answer

so

$$ 7(k+1)+1+(2k-6)\ge 7(k+1)+1$$

slaaibak · Answer

I would have done the proof a bit different.

$$k^2 + 2k + 1 > 7(k + 1) +1$$

$$(k^2) + (2k +1) > (7k + 1) + (7)$$

we know k^2 > 7k + 1

and we know 2k + 1 > 7 since k>=8

zarkon · Answer

I would have used calculus ;)

anonymous · Answer

@Zarkon I would too.

anonymous · Answer

Can we agree that the example given above is not good?