Mathematical Induction (The inductive step)

[pavel329]

1. I don't understand how to prove this.
for all n≥1, 10ⁿ - 1 is divisible by 9.





3. I've done the basis step.
Now I'm on the inductive step.
I'm using (10^k+1-1)/9=1.
I don't know where to go from there.
Using algebra just gets me down to 10^k+1= 10. And I really don't think that's the answer.
All examples I've seen show me things like 1...2..3..n+1= n(n+1) or something along the lines. They already give me the equation. This one does not.

 

[Mark44]

1. I don't understand how to prove this.
for all n≥1, 10ⁿ - 1 is divisible by 9.





3. I've done the basis step.
Now I'm on the inductive step.
I'm using (10^k+1-1)/9=1.

This isn't right. You need to show that 10^k+1-1 is divisible by 9, not that it is equal to 9. There is a difference. For example, 27 is divisible by 9, but the two numbers aren't equal.

I don't know where to go from there.
Using algebra just gets me down to 10^k+1= 10. And I really don't think that's the answer.
All examples I've seen show me things like 1...2..3..n+1= n(n+1) or something along the lines. They already give me the equation. This one does not.

What do you have for your induction hypothesis?

 

[ehild]

Assume that 10^k-1 is divisable by 9. Prove that it implies 10^k+1-1 is also divisable by 9.

Try to bring 10^k+1-1 to such form that it contain 10^k-1.
10^k+1=10*10^k=> 10^k+1-1=10*10^k-1=10*10^k-10+10-1=
10(10^k-1)+9.
Can you proceed from here?


ehild