MHB 1.1.21 simplify rational expression

AI Thread Summary
The discussion focuses on simplifying the rational expression $\left(\dfrac{a^2b^3-2a^{-3}b^3}{2a}\right)^2$. Participants agree that simplifying by canceling "a" before squaring is beneficial, leading to the expression $\frac{\left(ab^3-2a^{-4}b^3\right)^2}{4}$. They further factor out $b^3$ and square the remaining terms, resulting in $\frac{b^6(a^2-4a^{-3}+4a^{-8})}{4}$. A debate arises about the treatment of negative exponents and whether they constitute fractions, with differing opinions on how to represent them. The conversation highlights the complexity of defining "simplification" in mathematical expressions.
karush
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simplify
$\left(\dfrac{a^2b^3-2a^{-3}b^3}{2a}\right)^2=$
OK this could get confusing quickly
but I don't think we want to square it first
sscwt.png
 
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Yes, it can be simplified by cancelling "a" before squaring.
(The "2" in the denominator is already simple.)
$\frac{\left(ab^3-2a^{-4}b^3\right)^2}{4}$

Of course we can also factor out that $b^3$ that is in each term. Squaring it gives $b^6$.

$\frac{b^6(a- 2a^{-4})^2}{4}$

Now it is relatively easy to square $a- 2a^{-4}$.
$(a- 2a^{-4})^2= a^2- 2(a)(2a^{-4})+ (2a^{-4})^2$
$= a^2- 4a^{-3}+ 4a^{-8}$

$\frac{b^6(a^2-4a^{-3}+4a^{-8}}{4}$.
 
ok we still have negative exponents in the numerator??
 
Is that a problem? Different people have different ideas of what "simple" is. If you don't want negative exponents, remember what a negative exponent means:
$\frac{b^2- 4a^{-3}+ 4a^{-8}}{4}= \frac{b^2- 4\frac{1}{a^{3}}+ 4\frac{1}{a^{8}}}{4}$

Now, to eliminate the "fractions in fractions", multiply both numerator and denominator by $a^8$
$\frac{a^8b^3- 4a^5+ 4}{4a^8}$
 
well the few examples I saw they brought the terms with negative exponents in the numerator to the denumerator
 
That is exactly what I did!
 
Country Boy said:
That is exactly what I did!

yes but you created fractions over over a fraction $ \dfrac{a^{-3}}{4}=\dfrac{1}{4a^3}$

anyway,, just being cranky
 
No, a^{-3} is not a fraction!
 
why not the exponent is positive
 
  • #10
Country Boy said:
No, a^{-3} is not a fraction!

It is when it's written as $\displaystyle \begin{align*} \frac{1}{a^3} \end{align*}$...
 
  • #11
Yes, $\frac{1}{x^3}$ is a fraction. $x^{-3}$ is NOT.
It is a matter of the difference in numerals, not numbers.
 
  • #12
numerals?
 
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  • #13
Country Boy said:
Yes, $\frac{1}{x^3}$ is a fraction. $x^{-3}$ is NOT.
It is a matter of the difference in numerals, not numbers.

Do you realize that you are saying that two amounts that are equivalent are simultaneously a fraction and not a fraction?
 
  • #14
they just said move the terms with negative exponents in the numerator to the denominator
the bell rang and I left
no fancy stuff
 
  • #15
Prove It said:
Do you realize that you are saying that two amounts that are equivalent are simultaneously a fraction and not a fraction?
Yes, I am. They are two different numerals that repreent the same number. The term "fraction" has to do with the way a number is represented, it is not a property of the number itself.
 

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