# Inequality Truths, am I doing this right?

• Sep 4th 2009, 02:25 PM
A Beautiful Mind
Inequality Truths, am I doing this right?
3. If 2 < x < 6, which of the following statements about x are necessarily true, and which are not necessarily true?

a) 0 < x < 4

b) 0 < x-2 < 4

c) 1 < x/2 < 3

d) 1/6 < 1/x < 1/2

e) 1 < 6/x < 3

I know what they're asking. They're asking which ones are true and which ones aren't. But the 2 < x < 6?

The answer is not necessarily true (saw it in the back of the textbook but there aren't any explanations as to why this is so).

Is it because:

$\displaystyle 2 < x < 6$
$\displaystyle 0 < x < 4$

Well it would be x > 2, so x > 0 would be true. And then the next part x < 6 but x < 4 could not be true because the answer could be 4 or 5?

And for this one:

$\displaystyle 2 < x < 6$
$\displaystyle 0 < x -2 < 4$

x > 2
x > 0

Well that wold be true because if x > 2, then x is definitely greater than 0. And if x <6 and x-2 < 4 would mean that x would have to be at least 5 since x is less than 6 and if you use 5-2 = 3 <4, that holds true.

And then this one:
$\displaystyle 2 < x < 6$
$\displaystyle 1 < x/2 < 3$

x is greater than 2, so 1/2x is greater than 1. Well I think that's true because if you have x greater than 2 then you can try 3 immediately and half of 3 is 1.5, which is greater than one. And if x <6, then the highest you can half is 5 and 1/2(5) = 2.5, which is less than three. This would be true.

$\displaystyle 2 < x < 6$
$\displaystyle 1/6 < 1/x < 1/2$

I don't know how to approach this problem now. I'll try though...

If x > 2 then on the bottom we can put in 1/3... and 1/3 is greater than 1/6th. 1/3 would be less than 1/2. Plug in 4... 1/4 is greater than 1/6 and 1/4 is less than 1/2. And since we're restricted to x < 6, the last value we plug in will be 5 to see if this is true...

1/6 < 1/5< 1/2.

1/5>1/6 and 1/5 < 1/2. So this true.

$\displaystyle 2 < x < 6$
$\displaystyle 1 < 6/x < 3$

x > 2 so 6/x > 1. Plug in 3 = 6/3=2 > 1. 6/5 = 1.2 > 1...so this is true too.

Am I doing this right?

There's one here that I didn't write above but it's an absolute value problem.

So...
$\displaystyle 2 < x < 6$
$\displaystyle |x-4| < 2$

x > 2 which means 3-4 = -1 and the |-1| = -(-1) = 1 < 2.

Take it up x being 5 and |5-4| = 1 absolute value is, so 1 < 2. This would be true as well.

$\displaystyle 2 < x < 6$
$\displaystyle -6 < -x < 2$

Since x has to be greater than 2, we can plug a 3 in. That would make it -3 > -6. True. Now -3 < 2 is true. And from the original problem it has to be x > 2 and x>6. So we're only using the values of 3, 4, and 5.

Plug in 5.

-5>-6. True.

-5< 2. True.

This is true.

Am I doing this all right?
• Sep 5th 2009, 09:45 PM
Mozart
Hi, Beautiful Mind. I think you have done great!

a) $\displaystyle 0 < x < 4$. The original inequality $\displaystyle 2 < x < 6$ says that $\displaystyle x$ could be any number greater than $\displaystyle 2$ and less than $\displaystyle 6$; and that includes, for example, the number $\displaystyle 5$, which the $\displaystyle 0 < x < 4$ doesn't cover. On the other hand $\displaystyle 0 < x < 4$ says that $\displaystyle x$ could be any number between $\displaystyle 0$ and $\displaystyle 4$. This includes the numbers $\displaystyle 1$ and $\displaystyle 2$ which are not true for $\displaystyle 2 < x < 6$. Thus $\displaystyle 0 < x < 4$ is not necessarily true. It could possibly be true though, because both the inequalities can take the values $\displaystyle 3$ and $\displaystyle 4$, for example. But it is not necessarily true.

b) $\displaystyle 0 < x-2 < 4$. It says that when $\displaystyle x$, which is greater than $\displaystyle 2$ and less than $\displaystyle 6$, is substracted by $\displaystyle 2$, its value falls between $\displaystyle 0$ and $\displaystyle 4$. The original inequality was $\displaystyle 2 < x < 6$. We want to find the region for which holds the truth for $\displaystyle x-2$, when $\displaystyle x$is greater than $\displaystyle 2$ and less than $\displaystyle 6$. The result which we then get is the only the one which the inequality holds true for $\displaystyle x-2$. Thus if what we find doesn't turn out to be $\displaystyle 0 < x-2 < 4$, then the inequality is not necessarily true. If, on the other hand, we find $\displaystyle 0 < x-2 < 4$ to be the region for which holds for $\displaystyle x-2$, when $\displaystyle x$ is greater than $\displaystyle 2$ and less than $\displaystyle 6$, then the inequality is necessarily true. To proceed, since we need to find the inequality for $\displaystyle x-2$, it's necessary that we have to substract $\displaystyle 2$ from all the parts of the inequality; it's by rule that what is done to one side of an inequality should as well be done to the other. So we have $\displaystyle 2-2 < x-2 < 6-2$, which is gives $\displaystyle 0 < x-2 < 4$. Hence $\displaystyle 0 < x-2 < 4$ is necessarily true.

c) $\displaystyle 1 < x/2 < 3$. This says when a value of $\displaystyle x$ greater than $\displaystyle 2$ and less than $\displaystyle 6$ is divided by $\displaystyle 2$ the answer will fall between $\displaystyle 1$ and $\displaystyle 3$. Ignore this for a minute and remember our original inequality. It was $\displaystyle 2 < x < 6$, basically saying that $\displaystyle x$ is greater than $\displaystyle 2$ and less than $\displaystyle 6$. Now, you want to find the region for which the inequality is true $\displaystyle 2 < x < 6$ is true for when $\displaystyle x$ is divided by $\displaystyle 2$. Since, by rule, whatever you do to one side of an inequality or an equation should as well be done to the other, you divide all the sides of the inequality by $\displaystyle 2$. That is $\displaystyle 2/2 < x/2 < 6/2$ which give you $\displaystyle 1 < x/2 < 3$. Thus $\displaystyle 1 < x/2 < 3$ is necessarily true.

d) $\displaystyle 1/6 < 1/x < 1/2$. This says when you take the reciprocal of $\displaystyle x$ {when you divide $\displaystyle 1$ by $\displaystyle x$}, where $\displaystyle x$, as the original inequality holds, is greater than $\displaystyle 2$ and less than $\displaystyle 6$, the value of $\displaystyle x$ lies between $\displaystyle 1/6$ and $\displaystyle 1/2$. To check this, let us start with the original inequality. $\displaystyle 2 < x < 6$. Since we need to find the inequality which holds true for the reciprocal of $\displaystyle x$ {when $\displaystyle 1$ is divided by $\displaystyle x$}, and since by rule anything done to one side of an inequality should as well be done to the other, we have to take the reciprocal of the whole inequality. This is done by dividing 1 by the whole inequality. So we have $\displaystyle \frac {1}{2 < x < 6}$. Now, if you take the reciprocal of an inequality, the sign of the inequality reverses. Thus $\displaystyle \frac {1}{2 < x < 6} = \frac {1}{2} > \frac{1}{x} > \frac {6}{2} = 1/2 > 1/x > 3$, which is another way of writing $\displaystyle 1/6 < 1/x < 1/2$, since both mean that $\displaystyle 1/x$ lies between $\displaystyle 1/6$ and $\displaystyle 1/2$. Thus $\displaystyle 1/6 < 1/x < 1/2$ is necessarily true.

e) $\displaystyle 1 < 6/x < 3$. This is actually similar to the above. You take the same steps; divide the $\displaystyle 6$ by the original inequality, and then if the result that comes out is $\displaystyle 1 < 6/x < 3$, the inequality ($\displaystyle 1 < 6/x < 3$) is necessarily true; if that doesn't come up, then it isn't. So, let us proceed:

$\displaystyle \frac {6}{2 < x < 6} = \frac {6}{2}> \frac {6}{x}> \frac {6}{6} = 3 > 6/x > 1 = 1 < 6/x < 3$.

Therefore $\displaystyle 1 < 6/x < 3$ is necessarily true.