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I am trying to complete an induction proof on this problem, but cannot figure out the final steps. I understand the logic behind what makes this theorem true, and will explain it here before I dive into what I have written so far for the proof.

For example, take $2^4$. To check if this is a deficient number, we look at the sum of it's divisors, which are $2^0 + 2^1 + 2^2 + 2^3$.

$2^4 = 16$ and $2^0 + 2^1 + 2^2 + 2^3 = 15$. Since $16 > 15$, we can say that $2^4$ is a deficient number.

This can be generalized to $2^n > \sum_{i = 0}^{n-1} 2^{i}$

For my proof so far, this is what I have:

Proof (by induction): Let $n$ be any positive integer, where we need to show $2^n$ is deficient. Also let the sum of the divisors of $2^n$ equal $2^0 + 2^1 + ... + 2^{n-1}$. We can say that $2^n$ is deficient if it is greater than the sum of it's divisors. For the base case, let $n = 1$. Then, $2^n = 2$, and the sum of the divisors is equal to $2^0$. Since $2^1 > 2^0$, the theorem holds true for this case.

Next, we want to find if the theorem holds for $n + 1$. Therefore: ... This is where I need help

I want to try and use $2^{n+1} > \sum_{i = 0}^{n} 2^{i}$ in some way, but don't know how as it has been a while since I've worked on proofs. Any help or alternative ideas would be greatly appreciated!

Theo Bendit
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Pecker
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  • If you know the formula for the sum of a geometric progression, then you can avoid induction. – Intelligenti pauca Apr 13 '21 at 16:20
  • @Intelligentipauca How might I go across doing this? – Pecker Apr 13 '21 at 16:24
  • I recommend this answer whenever someone wants to structure an induction proof. You structure the sum for $n+1$ as the sum for $n$ plus whatever is new. Use your assumption for the sum for $n$ – Ross Millikan Apr 13 '21 at 16:24
  • Might be easier and be less likely to trip on your feet if you prove the stronger statement $\sum_{d|2^n} d = \sum_{k=0}^{n-1} 2^k = 2^n -1$. Then $\sum_{d|2^{n+1}} = \sum_{k=1}^n 2^k = (\sum_{k=0}^{n-1} 2^k) + 2^n= (2^n-1)+2^n = 2^{n+1} -1$.... Although you can simply say $\sum_{k=0}^{n-1} 2^k \ne 2^n$ so $\sum_{k=0}^n 2^k \ne 2^n + 2^n = 2^{n+1}$. But you are likely to trip on your feet if you do that. – fleablood Apr 13 '21 at 16:47

2 Answers2

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We have $$2^n\gt \sum_{i=0}^{n-1} 2^i$$ and so we have $$2^n\ge 1 + \sum_{i=0}^{n-1} 2^i$$ because both sides are integers, and so $$2^{n+1}\ge 2 + \sum_{i=1}^{n} 2^i $$ $$= 1 + \sum_{i=0}^{n} 2^i $$ $$\gt \sum_{i=0}^{n} 2^i$$.

JMP
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  • A few questions about this. What properties allow you to include $... \ge 1 + ...$? From my understanding, you are using $2^0$, and then when $n+1$ comes in, it becomes $2^1 = 2$. My next question is about the ending of the proof. Are you simply saying $2^{n+1} > \sum_{i=0}^{n} 2^i$? – Pecker Apr 13 '21 at 16:53
  • If $x,y\in\mathbb{Z}$ and $x\gt y$ we have $x\ge y+1$ and also $2x\gt 2y+1$. – JMP Apr 13 '21 at 16:54
  • So $2^n$ is $x$, and $\sum_{i=0}^{n-1} 2^i$ is $y$ in this scenario? – Pecker Apr 13 '21 at 16:59
  • A proof by induction assumes the proposition is true for $n$ and then uses some algebra to prove that the proposition is then true for $n+1$. As is it also true for the base case, it is now true for all $n$. I have assumed the result is true for $n$, and then using this, show that is would then be true for $n+1$ as well. – JMP Apr 13 '21 at 17:00
  • @Pecker; yes, and we know $x$ is strictly greater than $y$, and so we can also say without doubt that $x$ is greater than or equal to $y+1$ as they are both integers.. – JMP Apr 13 '21 at 17:01
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$1 + 2 + ...... + 2^{n-1} \ne 2^n$ so

$1 + 2 + ..... + 2^{n-1} + 2^n \ne 2^n + 2^n =2\cdot 2^n =2^{n+1}$

fleablood
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  • Could you elaborate on how this proves deficiency with $2^n$? – Pecker Apr 13 '21 at 16:51
  • Huh? That was your induction hypotheses. If $2^n$ was deficient this proves the deficiency of $2^{n+1}$. If you want to prove $2^n$ is deficient then your induction hypothesis is that $2^{n-1}$ is deficient. If $1 + 2 + ..... + 2^{n-2} \ne 2^{n-1}$ then $1+ 2+ .... + 2^{n-2} + 2^{n-1} \ne 2^{n-1} + 2^{n-1} = 2^n$. – fleablood Apr 13 '21 at 17:30
  • I see now. I thought you were saying something different with the $\ne$ sign. Thank you! – Pecker Apr 13 '21 at 17:33