Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the following two inequalities: $$\begin{align*} 8x &\gt 12y\\ 12y &\gt 15z \end{align*}$$ Now the book states that we need to line up the inequalities as such $$\begin{array}{rcccccl} 0 &<& 15z\\ && 15z &<& 12y\\ & & & & 12y &<& 8x\\ \end{array}$$ Hence we get $$ 0 < 15z < 12y < 8x $$

Now my question is how did the book get the following $$\begin{array}{rcccccl} 0 &<& 15z\\ && 15z &<& 12y\\ && && 12y &<& 8x \end{array}$$

share|improve this question
    
Thats what i want to know how did the book get $$ 0<15z $$ , I could only come up with $$ 15z < 12y < 8x $$ –  Rajeshwar Jul 9 '12 at 23:05
    
The $0\lt 15z$ part is not sensible, since nothing in the two given inequalities implies that $0 \lt 15z$. As to the others, $p \gt q$ says the same thing as $q \lt p$. So for example the given inequality $8x \gt 12y$ can be rewritten as $12y \lt 8x$. –  André Nicolas Jul 9 '12 at 23:06

2 Answers 2

up vote 0 down vote accepted

You know that $12y\lt 8x$, because that is the first inequality you have; it appears last in the large display.

You also know that $15z\lt 12y$, because that is the second inequality you have; it appears in the middle of the large display.

And, presumably, you know that $z$ is positive, so that $15z$ is also positive, $15z\gt 0$.

So: $0$ is smaller than $15z$; and $15z$ is smaller than $12y$; and $12y$ is smaller than $8x$. That's the three inequalities that appear, only they are indented to make it clear how they fit together.

share|improve this answer

When you write a compound inequality such as $1 < x < 2$, you are using a tacit Boolean and. So, for example, $$ (1 < x < 2) \Leftrightarrow ((1 < x) \wedge (x < 2)).$$ This is how you should parse it.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.