How does $1.88x < z <2x$ imply $z>6$? 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 $z>6$?
 A: 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 $z \ge 17$.
A: 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.
A: 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 $2x$. 
A: 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 $6$.
A: 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.
A: 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?
