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Can we demonstrate this fact: that the bigger the numbers are the more cubes are between $10^n$ and $2*10^n$?

I need this to solve a problem. Here is my idea.

I think we have to show that if $10^x≤a^3≤2∗10^x$, $10^y≤b^3≤2∗10^y$ and $x<y$, then b has more solutions than a.

We can write the 2 inequality as $\sqrt[\leftroot{10} \uproot{5} 3]{10^x} \leq a \leq \sqrt[\leftroot{10} \uproot{5} 3]{10^x*2}$

$\sqrt[\leftroot{10} \uproot{5} 3]{10^y} \leq b \leq \sqrt[\leftroot{10} \uproot{5} 3]{10^y*2}$

which basically means that

$\sqrt[\leftroot{10} \uproot{5} 3]{2*10^x} -\sqrt[\leftroot{10} \uproot{5} 3]{10^x} \leq \sqrt[\leftroot{10} \uproot{5} 3]{10^y*2} -\sqrt[\leftroot{10} \uproot{5} 3]{10^y}$

$\sqrt[\leftroot{10} \uproot{5} 3]{10^x} (\sqrt[\leftroot{10} \uproot{5} 3]{2}-1) \leq \sqrt[\leftroot{10} \uproot{5} 3]{10^y}(\sqrt[\leftroot{10} \uproot{5} 3]{2}-1) $

$\sqrt[\leftroot{10} \uproot{5} 3]{10^x} \leq \sqrt[\leftroot{10} \uproot{5} 3]{10^y}$, which is true because $x<y$.

Is my idea right?

Hope one of you can help me! Thank you!

KDP
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IONELA BUCIU
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  • Your approach is good. Obviously, you also can notice that number of integer $a$ such that $10^n \le a^3 \le 2 \cdot 10^n$ is non-decreasing function of $n$. – openspace Nov 21 '23 at 08:59
  • Why is there a downvote? – Mr.Gandalf Sauron Nov 21 '23 at 09:39
  • You have shown a relationship between x and y. Not a and b. – Starlight Nov 21 '23 at 09:48
  • @Starlight Yeah, but that relation is $\sqrt[\leftroot{10} \uproot{5} 3]{10^x} -\sqrt[\leftroot{10} \uproot{5} 3]{10^x2} \leq \sqrt[\leftroot{10} \uproot{5} 3]{10^y} -\sqrt[\leftroot{10} \uproot{5} 3]{10^y2}$, which basically makes b have more solutions. – IONELA BUCIU Nov 21 '23 at 10:17
  • You have things in reverse. $\sqrt[3]{10^x} < \sqrt[3]{2\cdot 10^x}$, so $\sqrt[3]{10^x} -\sqrt[3]{2\cdot 10^x} < 0$. The larger the magnitude of a negative number, the lower the number will be. Thus $$\sqrt[3]{10^y} -\sqrt[3]{2\cdot 10^y} < \sqrt[3]{10^x} -\sqrt[3]{2\cdot 10^x}$$. – Paul Sinclair Nov 22 '23 at 05:26
  • @PaulSinclair I edited the mistype. is my idea correct now? – IONELA BUCIU Nov 22 '23 at 07:12
  • @mathlove I edited the mistype. is my idea correct now? – IONELA BUCIU Nov 22 '23 at 07:13
  • Just to be clear, are you saying that if we have $a\le x\le b, c\le y\le d$ and $b-a\lt d-c$ where $a,b,c,d$ are real numbers, then the number of such integer $y$ is larger than the number of such integer $x$ ? – mathlove Nov 22 '23 at 07:29
  • @mathlove Yes. This is what I'm saying. – IONELA BUCIU Nov 22 '23 at 07:43
  • Then, what you are saying is not always true. There is a counterexample. Take $a=0.9,b=2.1,c=1.1$ and $d=2.9$. We have $b-a\lt d-c$, but the number of such integer $x$ is $2$ while the number of such integer $y$ is $1$. – mathlove Nov 22 '23 at 07:44
  • @mathlove Yes, but $a,c$ take the form of $10^x$ and $b,d$ take the form $2*10^x$ and all of them are natural. The idea is still wrong now? – IONELA BUCIU Nov 22 '23 at 07:50
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    Note that $a,c$ take the form of $\sqrt[3]{10^x}$ and $b,d$ take the form $\sqrt[3]{2*10^x}$ and all of them are not always natural. I would say that your idea does not work. – mathlove Nov 22 '23 at 08:09
  • Related to this question by the OP: https://math.stackexchange.com/q/4807359 – PM 2Ring Nov 22 '23 at 09:05
  • The problem with your approach is that you have mixed up "scratchwork" used to figure out how to demonstrate something with the demonstation itself. To actually show something is true, you start with what you already know is true, and derive from it what you want to show. But in your post, you started with what you wanted to show and derived from it something true. You need to reverse your logic: $$x < y \implies 10^{x/3} < 10^{y/3} \implies 10^{x/3}(\sqrt[3]2-1) < 10^{y/3}(\sqrt[3]2-1)\\implies \sqrt[3]{2\cdot 10^x} - \sqrt[3]{10^x} < \sqrt[3]{2\cdot 10^y} - \sqrt[3]{10^y}$$ – Paul Sinclair Nov 22 '23 at 14:19
  • Then as per gnasher729's post, note that $y > 1$ is required to assure that there is an integer between $\sqrt[3]{10^y}$ and $\sqrt[3]{2\cdot 10^y}$. That is, that $\sqrt[3]{2\cdot 10^y} - \sqrt[3]{10^y} > 1$. – Paul Sinclair Nov 22 '23 at 14:21

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Close, but false for n = 0 and n = 1. There is ONE cube from 1 to 2 but no cube from 10 to 20.

The number of cubes can be slightly greater than the difference of the two cube roots, or slightly less. So n must not be very small.

gnasher729
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