Prove by induction: $10^0 + 10^1 + 10^2 + ... + 10^n < 10^{n+1}$, $n >= 0$.

We have the predicate $$P(n) = 10^0 + 10^1 + 10^2 + ... + 10^n < 10^{n+1}$$, $$n >= 0$$

To prove by induction we need to check the base case $$P(0)$$, and after that we have our induction hypotheses where we assume that the predicate is true for some $$k = n$$. After that we need to show that if $$P(k)-->P(k+1)$$ then it's true for all $$n>=0$$.

Base case: $$P(0) = 10^0 = 1 < 10^1 = 10$$ TRUE

Induction hypothesis: $$10^0 + 10^1 + 10^2 + ... + 10^k < 10^{k+1}$$, for some $$k = n$$.

Induction step $$P(k+1)$$: try to prove this

$$10^0 + 10^1 + 10^2 + ... + 10^k + 10^{k+1} < 10^{n+2}$$

...

That's where I'm lost. How can I show that $$10^0 + 10^1 + 10^2 + ... + 10^k + 10^{k+1}$$ is less than it's successor?

... I think I solved it. Thank you abiessu.

$$10^0 + 10^1 + 10^2 + ... + 10^k + 10^{k+1} < 10^{k+1} + 10^{k+1}$$

Know I need to show that $$10^{k+1} + 10^{k+1}$$ < $$10^{k+2}$$

$$10^{k+1} + 10^{k+1} = 2*10^{k+1}$$

$$10^{k+2} = 10*10^{k+2}$$

So we show that $$2*10^{k+1} < 10*10^{k+2}$$

$$10^0 + 10^1 + 10^2 + ... + 10^k + 10^{k+1} < 10^{k+1} + 10^{k+1} = 2*10^{k+1} < 10^{k+2} = 10*10^{k+2}$$

Therefore we can conclude our proof by mathematical induction that $$10^0 + 10^1 + 10^2 + ... + 10^k + 10^{k+1} < 10^{k+2} = 10*10^{k+2}$$

• By the inductive hypothesis it is less than $10^{k+1}+10^{k+1}$ Dec 14, 2022 at 14:44
• If $a\lt b$ with both integer, then $10a +1\lt 10b$ Dec 14, 2022 at 14:44
• Pick one of the two generic approaches described in this answer. Dec 14, 2022 at 14:48
• Just a slight remark : An easier proof is $11\cdots 11<100 \cdots 00$ where on the left side we have $n+1$ ones and on the right side $n+1$ zeros. Dec 14, 2022 at 14:48
• Your idea is correct, but you kept using $k+2$ in a couple places where you should use $k+1$. Dec 14, 2022 at 16:54

Your hypothesis is that $$10^0 + 10^1 + 10^2 + ... + 10^k < 10^{k+1}$$ Therefore $$10^0 + 10^1 + 10^2 + ... + 10^k + 10^{k+1} < 10^{k+1} + 10^{k+1}$$

Can you continue from there?

• Why did you modify $10^{k+2}$ by $10^{k+1} + 10^{k+1}$ ? $10^{k+2}$ is not equal to $10^{k+1} + 10^{k+1}$, I do not get it, can you explain it?
– Will
Dec 14, 2022 at 14:48
• @Will: what is another way to write $10^{k+2}$ in terms of $10^{k+1}$? Or, can you use $2\lt 10$? Dec 14, 2022 at 14:50
• @Will I didn't replace $10^{k+2}$, I just took the Induction hypothesis and add $10^{k+1}$ on both side of the inequality. From there, and using abiessu comment, you have to make the $10^{k+2}$ appear Dec 14, 2022 at 14:55
• @abissu but $10^{k+2} = 10^{k+1}.10^1$? How did you manipulate $10^{k+2}$ to become $10^{k+1} + 10^{k+1}$ ?
– Will
Dec 14, 2022 at 14:56
• @Will He did not modify $10^{k+2}$ in that way. All he has done to get from the first equation to the second equation is add $\ 10^{k+1}$ to both sides. Dec 14, 2022 at 15:07