# Induction

#### dunsta

Is the the correct approach to work through this problem

23n – 1 is divisible by 7 for all n>=1
n(1)
23(1) – 1 = (2*2*2) – 1
= 7

23n – 1 = 7m
n(k+1)
23(k+1) – 1 = 2.23k-1
= 2 * (7m+1) – 1
= 14m + 2 -1
= 14m-1
=7(2m-1)
Where (2m-1) is an int.
So 23n – 1 is divisible by 7.
Therefore, by P.M.I 23n – 1 is divisible by 7 for n>=1.

#### Prove It

MHF Helper
First, this is extremely difficult to read. Please learn some LaTeX.

You are trying to prove $$\displaystyle 2^{3n} - 1$$ is divisible by $$\displaystyle 7$$ for all $$\displaystyle n \geq 1$$ (I assume $$\displaystyle n$$ is also an integer)...

Base Step: $$\displaystyle n = 1$$.

Then $$\displaystyle 2^{3\cdot 1} - 1 = 8 - 1$$

$$\displaystyle =7$$ which is divisible by $$\displaystyle 7$$.

Inductive Step: Assume this statement is true for $$\displaystyle n = k$$.

Then $$\displaystyle 2^{3k} - 1 = 7m$$ where $$\displaystyle m$$ is an integer.

Now let $$\displaystyle n = k + 1$$ we have

$$\displaystyle 2^{3(k + 1)} - 1 = 2^{3k + 3} - 1$$

$$\displaystyle = 2^{3k}\cdot 2^3 - 1$$

$$\displaystyle = 8\cdot 2^{3k} + 8 - 7$$

$$\displaystyle = 8(2^{3k} + 1) - 7$$

$$\displaystyle = 8\cdot 7m - 7$$

$$\displaystyle = 7(8m - 1)$$

which is divisible by $$\displaystyle 7$$.

Therefore $$\displaystyle 2^{3n}- 1$$ is divisible by $$\displaystyle 7$$ for all integer $$\displaystyle n \geq 1$$.

dunsta

#### dunsta

Thanks ProveIt, I completely forgot about Latex for a minute and just pasted in the equation I was working on in word, so sorry!

How does the negative 1 in this line $$\displaystyle = 2^{3k}\cdot 2^3 - 1$$

change to negative 7?

$$\displaystyle =8\cdot 2^{3k} + 8 - 7$$

#### Prove It

MHF Helper
Sorry, that should actually read

$$\displaystyle 8\cdot 2^{3k} - 8 + 7$$

$$\displaystyle = 8(2^{3k} - 1) + 7$$.

The rest follows.

#### dunsta

ok, thanks
But how/where does the seven come from?

#### Prove It

MHF Helper
$$\displaystyle -8 + 7 = -1$$.

It just takes a bit of practice, I knew I needed to create a $$\displaystyle 7$$ in order to have this value be divisible by $$\displaystyle 7$$.

#### dunsta

Please excuse my ignorance, as mathematical Induction and I just aren't getting along.
But did you just place the "7" in the equation because you needed it. It didn't come from any other part of the equation?

#### Prove It

MHF Helper
That's correct. And I also realised I'd need to take out a common factor of $$\displaystyle 8$$.

Like I said, it just comes from experience.

dunsta
Similar threads