prove n < 10^n by induction

[pbem]

Hi, I'm also trying to solve this problem and I'm a little confused.

From last year's thread:

...prove n < 10^n.

for n=1, n<10^n because 1<10.
assume for n=k, k<10^k so n<10^n
therefore, k+1<10^k+1 (if a<b, a+c<b+c for any c)

for any positive integer k, 10^k will be a positive integer.
therefore, 10^(k+1)=10(10^k)>(10^k)+1....

You are very close...

As you pointed out 10ⁿ⁺¹ = 10 10ⁿ, so let's write that as
10ⁿ⁺¹ = 10 10ⁿ = (1 + 9) 10ⁿ = 10ⁿ + 9 10ⁿ > n + 9 10ⁿ
since 10ⁿ > n. Now, 10ⁿ > 1 for n>1 since 10^k > k > 1, k = 2, 3, 4, ..., n.

How is 10^n + 9(10^n) > n + 9(10^n) related to the previous step?

(The previous step being 10^n + 9(10^n) > n + 1)

I think I understand that 10^n + 9(10^n) > n + 9(10^n) is used so that like terms can cancel out (and that the induction hypothesis is used to show that 10^n > n, but I'm so confused how that step was made. Help would be appreciated, thank you!

 

[Ishuda]

Hi, I'm also trying to solve this problem and I'm a little confused.

From last year's thread:


How is 10^n + 9(10^n) > n + 9(10^n) related to the previous step?

(The previous step being 10^n + 9(10^n) > n + 1)

I think I understand that 10^n + 9(10^n) > n + 9(10^n) is used so that like terms can cancel out (and that the induction hypothesis is used to show that 10^n > n, but I'm so confused how that step was made. Help would be appreciated, thank you!

In essence what I was trying to say was
10ⁿ > n by induction assumption
9 10ⁿ > 9 n (by induction assumption) > 1
so
10ⁿ⁺¹ = 10ⁿ + 9 10ⁿ > n + 1