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If $z, x$ are positive integers, does $1.88 x< z <2x$ impliy $z>6$?

I found this in a paper. I notice $x > 8$ for $z$ to be integer.

How does $1.88x < z <2x$ imply $z>6$?

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    $\begingroup$ Start with small values of $x$. Can you find an integer in the open interval $(1.88, 2)$? In $(3.76, 4)$? Then you should see the idea (and find the solution quickly). $\endgroup$ – Martin R Jul 16 '19 at 12:22
  • $\begingroup$ That is an easy elementary approach, but a bit brute-forceish and doesn't extend well to a larger problem. What if it was $1.9999999x<z<2x$ instead and the claim was that $z>10000$... I wouldn't want to check each interval manually one at a time until I rule out all possibilities. $\endgroup$ – JMoravitz Jul 16 '19 at 12:24
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    $\begingroup$ @JMoravitz: Emphasis on “see the idea” :) $\endgroup$ – Martin R Jul 16 '19 at 12:26
  • $\begingroup$ "I notice x>8 for z to be integer" and "How does 1.88x<z<2x imply z>6?" The answer is because "I notice x>8 for z to be integer". $z > 1.88x > 1.88\cdot 8 = 15.04 > 6$ w $\endgroup$ – fleablood Jul 20 '19 at 0:03
  • $\begingroup$ Okay... I have to ask why you are asking if $z> 6$. We've proven $z> 15$ so why was $z > 6$ asked? Was that a necessary condition for something else that needed to be shown? $\endgroup$ – fleablood Jul 20 '19 at 0:22
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EDITED: Let $y = 2 x - z$. Then $y$ is an integer, and $y > 0$ so $y \ge 1$. But $y < 2 x - 1.88 x = 0.12 x$ so $x > y/0.12 = 25 y/3 \ge 25/3$, and that says $x \ge \lceil 25/3\rceil = 9$. And then $z > 1.88 x \ge 16.92$ so $z \ge 17$.

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    $\begingroup$ Shouldn't it be $y<2x-1.88x=0.12x$? $\endgroup$ – J_P Jul 16 '19 at 12:21
  • $\begingroup$ Thanks for catching. Edited. $\endgroup$ – Robert Israel Jul 16 '19 at 12:26
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For $x$ integer also $2x$ is integer. The closest integer value smaller than $2x$ is $(2x-1)$ - but that must be greater than $1.88x$ for $z$ to fit required boundings.

So: $$1.88x < 2x-1$$ which implies $$x>\frac 1{2-1.88} = \frac 1{0.12}$$ Then $1.88x > \frac {1.88}{0.12} > 15,$ so $z>15,$ which implies $z>6$,
Q.E.D.

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You just have to show that $x$ must be at least $4$ so that $z >4(1.88)>6$. For this note that when $x=1,2$ or $3$ there is no integer between $(1.88)x$ and $2x$.

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  • $\begingroup$ Your bounding $x\ge 4$ seems too weak. For $x=6$ there's no integer between $1.88\cdot 6=11.28$ and $2\cdot 6=12$, either! $\endgroup$ – CiaPan Jul 16 '19 at 12:38
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    $\begingroup$ True but I was only trying to prove what OP wanted. I could have gone further and found the smallest possible value of $z$. $\endgroup$ – Kavi Rama Murthy Jul 16 '19 at 12:41
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You've already essentially answered your own question, although the lower bound you give for $z$ seems somewhat arbitrary. You've already realised that $x \geq 9$, so you can easily substitute $9$ into the original inequality to obtain the lower bound for $x$, in this case $16.92 < z < 18 $, and the only integer solution of this is clearly $17$, which is strictly greater than $6$.

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Well, to be different.

$1.88 x < z < 2x$

$1.8333333.... x= 1\frac 56 x< z < 2x$

$11x < 6z < 12x$. Now as $3|6z$ and $3|12x$ inorder for $11x$ which is smaller then $12x$ to also be divisible by $3$ we must have $12x - 11x=x$ be a multiple of $3$.

If $x = 3$ then $33 < 6z < 36$ and $5.5 < z < 6$ which is impossible.

So $11x$ is a multiple of $3$ that is larger than $33$ so $11x \ge 36$ so...

$36 \le 11x < 6z$.

So $z > 6$. Which is exactly what you wanted!

But notice.....

This is a STUPID answer!

But I wonder if somehow something like this is why the value of $z>6$ rather than $z>15$ (which can also be proven) came up.

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Since $x\ge 1$, then $1.88x<z$ implies $z\ge 2$.

Then $z<2x$ implies $x\ge 2$.

Then $1.88x<z$ implies $z\ge 4$.

Then $z<2x$ implies $x\ge 3$.

Then $1.88x<z$ implies $z\ge 6$.

Then $z<2x$ implies $x\ge 4$.

Then $1.88x<z$ implies $z\ge 8$.

Can you continue?

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