# Prove that if $n \cdot 2^{-t} <0.01$ then $n \cdot 2^{-t} <\frac{1}{101}$

Is the following theorem true?

If $n \cdot 2^{-t} <0.01$ then $n \cdot 2^{-t} <\frac{1}{101}$ for $t,n \in \mathbb{N}$.

I've tried basic induction but that has led me nowhere, same with thinking of a counter-example.

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Is this not just the question wether $0.01\leq\frac{1}{101}$? –  drhab Oct 13 '13 at 14:29
$n=163$, $t=14$. (The set of numbers of the form $n\cdot 2^{-t}$ is dense in $\Bbb R$.) –  David Mitra Oct 13 '13 at 14:32
But that's false, isn't it? I want to know if you can put \frac{1}{101} between a multiple of a negative power of $2$ if you know that product is less than $0.01$. –  Arek Krawczyk Oct 13 '13 at 14:33

Let's find t such that if $2^{-t}$ is fixed and $n$ is increased one by one from $0$, $n2^{-t}$ will satisfy the inequality $1/101<n2^{-t}<1/100$. If such t and n exist, the theorem is clearly not true. It is obvious that if $n$ is increased one by one, $n2^{-t}$ satisfies the above inequality at some value of $n$ if $2^{-t}$ is smaller than

$1/100-1/101=1/10100$.

Since $2^{-13}>1/10100>2^{-14}$, if $t=14,$ there must be some $n$ such that $1/101<n2^{-t}<1/100$. This is enough as proof to disprove the theorem.

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For example, if $t=-14$, $n=163$ satisfies the above inequality, as the above person stated. –  Aran Komatsuzaki Oct 13 '13 at 14:54

1/100 > 1/101

Since, your equation is < 1/100, it can be <= 1/101

your n⋅2−t <= 1/101 < 1/100

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OP was asking if one could find a number of the form $n2^{-t}$ between $\frac 1{101}$ and $\frac 1{100}$. This does not answer the question. –  Ross Millikan Oct 13 '13 at 15:20

your statement isn't correct. What you say is if n⋅2^−t<0.01<1/101 and that can;t be true. Because 1/101 is always smaller than 0.01. So if n,t are smaller then 0.01 it doesn't automatticly mean that it is also smaller then 1/101

What is correct is, is: if n⋅2^−t<1/101 then n⋅2^−t<0.01

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OP was asking if one could find a number of the form $n2^{-t}$ between $\frac 1{101}$ and $\frac 1{100}$. This does not answer the question. –  Ross Millikan Oct 13 '13 at 15:21