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I am studying inequality and come across the following statement. I don't understand it and want to believe the book must have made mistakes. I am going to copy what the book says here exactly.

A linear inequality with one variable is in the form: $ax>b$. When $a>0$, the solution is $x>b/a$, When $a<0$, the solution is $x<b/a$. When $a=0$, if $b<0$, there are infinite number of solutions; if $b>0$, there are no solutions.

Asaf Karagila
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learning
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  • Can you provide more context? It's not clear what equation the "solutions" are to. Is it $ax-b=0$? – kccu May 27 '16 at 02:22
  • Are $a,b$ real numbers? – M47145 May 27 '16 at 02:25
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    @kccu I revised the question. Thank you for asking for details. – learning May 27 '16 at 02:28
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    @M47145 The book does not say whether a and b are real numbers, but I assume they are. – learning May 27 '16 at 02:29
  • In that case they presumably are. – M47145 May 27 '16 at 02:29
  • @M47145 I think the correct statement should be "When a=0, if b=0, there are infinite number of solutions; if b not equals to zero, there are no solutions." Am I right? If I am right, I have another question: Why if a=b=0, there are infinite number of solutions? What is reason behind it? – learning May 27 '16 at 02:37
  • @learning If $a=b=0$, then there also no solutions. This is for the same reason that there are no solutions if $a=0$ and $b>0$. I hope that helps. – M47145 May 27 '16 at 02:52

3 Answers3

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Your book is correct.

When $a>0$, multiply both sides of the inequality by $\frac{1}{a}$. Since this is a positive number, the direction of the inequality doesn't change: \begin{align*} ax&>b \\\frac{1}{a}\cdot ax &> \frac{1}{a} \cdot b\\ x&>\frac{b}{a} \end{align*}

When $a<0$, again multiply both sides by $\frac{1}{a}$. Now the direction of the inequality reverses since $\frac{1}{a}$ is negative. \begin{align*} ax&>b \\\frac{1}{a}\cdot ax &< \frac{1}{a} \cdot b\\ x&<\frac{b}{a} \end{align*}

When $a=0$, we cannot multiply by $\frac{1}{a}$. The inequality becomes $0\cdot x>b$, i.e., $0>b$. If $b>0$, there is a contradiction - $b$ cannot be both positive and negative. However, when $b<0$, every value of $x$ satisfies the equation $0 \cdot x >b$.

kccu
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If we have the inequality $a x > b$ then we can only divide by $a$ on both sides, if $a \ne 0$.

The next thing to watch out is that a negative value of $a$ will flip the $>$ into a $<$, while a positive $a$ will not change the operator, if we multiply or divide both sides by $a$.

This means $$ a x > b \wedge a > 0 \iff x > b / a $$ then $$ a x > b \wedge a < 0 \iff x < b / a $$ The third case was $$ a x > b \wedge a = 0 \quad (*) \iff 0 > b \wedge a = 0 \iff b < 0 \wedge a = 0 $$ so it depends on $b$ if there are infinite many solutions $x$ with statement $(*)$ true or not.

Your book is mostly correct, it only misses the case $a=b=0$, where there is also no solution $x$ which makes $(*)$ true.

mvw
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Your book is correct.

When $a=0$, our inequality is of the form $$0\cdot x-b>0.$$

If $b<0$, then $-b>0$. Now how many values of $x$ satisfy $0\cdot x -b>0$? No matter what we plug into $x$, we get that the inequality holds. Say $x=34$, then $0\cdot34-b=-b>0$. Since this is true no matter what we choose for $x$, there are infinitely many solutions.

If $b>0$, however, then $-b<0$. Now can you plug anything into $0\cdot x-b>0$ to make the statement true? No matter what you plug in for $x$, you get that $-b>0$, but this is false since we assumed that $-b<0$. So there is no solution $x$ that satisfies the inequality.

M47145
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