# 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 $$z>6$$?

• 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). – Martin R Jul 16 '19 at 12:22
• 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. – JMoravitz Jul 16 '19 at 12:24
• @JMoravitz: Emphasis on “see the idea” :) – Martin R Jul 16 '19 at 12:26
• "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 – fleablood Jul 20 '19 at 0:03
• 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? – fleablood Jul 20 '19 at 0:22

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 $$z \ge 17$$.

• Shouldn't it be $y<2x-1.88x=0.12x$? – J_P Jul 16 '19 at 12:21
• Thanks for catching. Edited. – Robert Israel Jul 16 '19 at 12:26

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.

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 $$2x$$.

• 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! – CiaPan Jul 16 '19 at 12:38
• True but I was only trying to prove what OP wanted. I could have gone further and found the smallest possible value of $z$. – Kavi Rama Murthy Jul 16 '19 at 12:41

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 $$6$$.

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.....

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.

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

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

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

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

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

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

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

Can you continue?