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Can you give me any suggestions?

I understand that it is the same as

$x\leq|({10^{10}})^{-j}|$

but I don't know how to conclude

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  • $\begingroup$ @coffeemath I already corrected it $\endgroup$
    – Kale_1729
    Sep 27, 2020 at 21:01
  • $\begingroup$ I saw that (have already erased comment). $\endgroup$
    – coffeemath
    Sep 27, 2020 at 21:04
  • $\begingroup$ @coffeemath Could you tell me how to do it? $\endgroup$
    – Kale_1729
    Sep 27, 2020 at 21:05
  • $\begingroup$ See my answer... let me know if more explanation needed. $\endgroup$
    – coffeemath
    Sep 27, 2020 at 21:11

3 Answers 3

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It depends on how you interpret the final phrase "with $j=1,2,3,\cdots.$" If it means just pick one of them then your answer is OK. If it means it should be true for every $k$ you need to intersect all the answers.

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  • $\begingroup$ Could be like this? $x\in \bigcap _{ j=1 }^{ \infty }{ ({ -10 }^{ -10j },{ 10 }^{ -10j }) } $ $\endgroup$
    – Kale_1729
    Sep 27, 2020 at 21:14
  • $\begingroup$ Yes, then finish by actually computing that intersection. $\endgroup$
    – coffeemath
    Sep 27, 2020 at 21:30
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The bigger the j is, the smaller your x should be. Therefore, if the question is for which values of x, $x≤|(10^{10})^{−j}|$ is always true no matter the j you're choosing, then the answer will be $0\leq x\leq 0$. You can prove this answer is correct by showing the limit of $|(10^{10})^{−j}|$ when j goes to $\infty$ is a number closer and closer to 0.

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  • $\begingroup$ I see, it's a very interesting suggestion, thank you very much. $\endgroup$
    – Kale_1729
    Sep 27, 2020 at 21:22
  • $\begingroup$ Can you give me a suggestion for this problem? $3^{-j}\leq x \leq 2^{-j}$ $\endgroup$
    – Kale_1729
    Sep 27, 2020 at 21:41
  • $\begingroup$ This one has no specific answer for all the j's because you can be tempted to choose $x = 0$ again by the same reasons than the previous one. However $\frac{1}{3}\leq 0 \leq \frac{1}{2}$ is false. And if you choose something bigger than 0, the limit when $j$ goes to $\infty$ is the following one: $0\leq x\leq 0$ which is false again. $\endgroup$
    – Figaro
    Sep 27, 2020 at 21:47
  • $\begingroup$ So how could it end? $\endgroup$
    – Kale_1729
    Sep 27, 2020 at 21:48
  • $\begingroup$ Could you conclude by saying: not true for any value of x? $\endgroup$
    – Kale_1729
    Sep 27, 2020 at 21:49
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If we extend the examples $$1^{-1}=1\qquad0.1^{-1}=10\qquad 0.01^{-1}=100\qquad 0.001^{-1}=100\qquad 0.0001^{-1}=10000\qquad $$ we can see that $${10}^{-10}\le x\le 1$$ if and only if $j=1.\quad$

For $j=2\implies 10^{-5}\le x \le 1$ and the trend is not exact with j-values that do not divide 10 evenly bu it works out if $j|10$, then

$$10^{-\frac{10}{j}}\le x \le 1$$

$$10^{-5}=0.00001\quad\land\quad 0.00001^2=10000000000=10^{10}$$ $$10^{-2}=0.01\quad\land\quad 0.01^5=10000000000=10^{10}$$ $$10^{-1}=0.1\quad\land\quad 0.1^{10}=10000000000=10^{10}$$

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