Date: 06/11/2002 at 08:33:07
From: Yasmin Rahmat
Subject: Algebra>Proofing and Logarithms
I actually have two questions that I got for a math project.
Question 1: If
2^2 + 3^2 = 3^2 + 2^2
subtract 2*2*3 from both sides, and we have
2^2 - 2*2*3 + 3^2 = 3^2 -2*2*3 + 2^2
so in algebra we know that
a^2 - 2ab + b^2 = (a-b)^2
So,
(2-3)^2 = (3-2)^2
Next square root both sides:
sqrt(2-3)^2 = sqrt(3-2)^2
Hence,
2 - 3 = 3 - 2
-1 = 1
Question 2:
Consider
1/4 > 1/16
which is the same as
(1/2)^2 > (1/2)^4
Take the log of both sides,
log(1/2)^2 > log(1/2)^4
By the power law, this is
2 log 1/2 > 4 log 1/2
Divide both sides by log 1/2, and we have
2 > 4
Please try and help me! I'm totally stuck!
Thanks
Date: 06/11/2002 at 12:43:56
From: Doctor Peterson
Subject: Re: Algebra>Proofing and Logarithms
Hi, Yasmin.
You just have to look at each step and ask yourself exactly what it
is doing. For example, what is (2-3)^2? What is its square root? Is
it true that this is the same as 2-3? If not, why not? What rule is
being followed, and what is the correct rule?
Again, when you work with inequalities, you have to multiply or
divide only by positive numbers, or you have to reverse the
inequality. Check every number you multiply by: is it positive? Don't
just assume that it is because there is no minus sign showing!
The purpose of this kind of assignment is to make it clear to you
that you must always think when you do algebra; it's easy just to let
the algebra carry you lazily along, without ever looking closely
enough to see what it is actually doing, and that can lead to big
mistakes. The same is true in life!
If you have any further questions, feel free to write back.
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/