Raising Both Sides of an Inequality to a Power
Date: 06/19/2006 at 05:52:52 From: Antti Subject: When to turn < or > around when raising to nth power First I have to say your website is marvelous. Thank you for that. I have one question about inequalities. You know when -2 < 1 and both sides are multiplied by -1, the < has to be changed to > to get the right answer, 2 > -1. What if we raise both sides of -2 < 1 to the 2nd power? Then we also have to turn the < around to get the right answer (-2)^2 > 1^2 = 4 > 1. How do I know when the < or > has to be turned around when raising to powers? Is there a rule for raising to the 3rd, 4th, or nth power?
Date: 06/19/2006 at 10:18:41 From: Doctor Peterson Subject: Re: When to turn < or > around when raising to nth power Hi, Antti. The inequality has to be turned around when you multiply by a negative number, since that essentially flips the whole number line around, reversing the order of everything. But squaring or raising to a higher power doesn't work that way; you can't always obtain an equivalent inequality at all. For example, suppose you know that a < b. Perhaps a is -2 and b is 1; or perhaps a is -1 and b is 2. What happens when we square each of these numbers? -2 < 1 becomes 4 > 1; the square of -2 is GREATER than the square of 1 -1 < 2 becomes 1 < 4; the square of -1 is LESS than the square of 2 So whether the inequality is reversed depends on the specific numbers you have, in particular on their absolute values. We can't say something like if a < b, then a^2 < b^2 All we can say is if |a| < |b|, then a^2 < b^2 So your question is in a sense invalid: yes, we can determine whether the square of this is less than the square of that, but the result is not derived in any way from the given inequality. We aren't really turning anything around at all. Now, with an ODD power, we CAN say, for example, that if a < b, then a^3 < b^3 You can see this if you look at the graph of y=x^3. The graph is always increasing (rising to the right), so if you pick two numbers a and b, with a < b, then the cube of the number on the right, b, is greater than the cube of the number on the left, a. That doesn't work with squaring (or any even power). Look at the graph of y=x^2; it falls for negative x, then rises for positive x. So if you pick a and b with a < b, you have no guarantee that a^2 < b^2 or not. The lesson for you: when you're working with inequalities, DON'T raise both sides to an even power. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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