Understanding Rational Inequalities: Why Can't We Multiply by the Denominator?

[rambo5330]

Homework Statement

Quick question on Rational Inequalities..

Say I have

\frac{1+x}{1-x} \huge\geq 1

Why can we not multiply 1 by denominator (1-x) is this because if x > 1 then (1-x) would be negative in effect changing the sign of the inequality.. but if x<1 then the inequality is preserved? therefore we subtract 1 from either side..setting the inequality equal to zero?

sorry the formatting isn't quite right on that expression but the left hand side is all one expression .. greater than or = 1

 

[rock.freak667]

Why can we not multiply 1 by denominator (1-x) is this because if x > 1 then (1-x) would be negative in effect changing the sign of the inequality.. but if x<1 then the inequality is preserved?

Yes.

therefore we subtract 1 from either side..setting the inequality equal to zero?

You can do this as well.

Alternatively, you can multiply by (1-x)².

 

[HallsofIvy]

What you can do is think "IF 1- x> 0, then I can multiply both sides by it and get 1+ x\ge 1- x". Add x to both sides and subtract 1 from both sides and you have 2x\ge 0 which is the same as x\ge 0. Now that was assuming 1- x> 0 which is the same as 1> x. So you have 0\le x&lt; 1.

Now, IF 1- x< 0, multiplying both sides by a negative number reverses the inequality: 1+ x\le 1- x. Again, add x to both sides and subtract 1 from both sides and you have 2x\le 0 or x\le 0. The "IF" is true as long as 1< x which can't be true if x\le 0 so this gives NO new solution.

The algebra gets a bit complicated but, yes, you can subtract 1 from each side: \frac{x+1}{1-x}- 1= \frac{x+ 1- (1- x)}{1-x}= \frac{2x}{1-x}\ge 0 which is true as long as the numerator and denominator have the same sign.

That is, both 2x> 0 and 1- x> 0 which gives x> 0 and x< 1 as before, or both 2x< 0 and 1- x< 0 which gives x< 0 and x> 1 which is impossible.

 

[rambo5330]

thanks for the excellent explanation!