Quick Math Question: Simplifying Fractions and Roots

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Homework Help Overview

The discussion revolves around simplifying mathematical expressions involving fractions and roots, specifically the equivalence of expressions and the simplification of square roots.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore whether the expression 1 / (ab/d)^x/y is equivalent to (d/ab) ^x/y and discuss the implications of order in such expressions.
  • There are attempts to simplify the expression root 10/root 160 and to clarify the correct surd form.
  • Questions arise regarding the application of exponents to fractions and the general case for roots of negative numbers.

Discussion Status

Some participants express agreement on the equivalence of the expressions and provide reasoning based on arithmetic rules. Others seek clarification on the application of exponents and the simplification process, indicating an ongoing exploration of the concepts involved.

Contextual Notes

Participants are navigating the rules of arithmetic and exponentiation, with some uncertainty about the application of these rules in specific contexts. There is also mention of potential variations in expressing roots.

roger
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Hello,

please can someone help me with this :

Is 1 / (ab/d)^x/y the same as (d/ab) ^x/y ?

Does the order matter ?



Also, if I have to simplify root 10/root 160 and put it into surd form is 1/4
wrong ?


thanks


Roger
 
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[tex]\frac{1}{(ab/d)^{x/y}}=(\frac{1}{(ab/d)})^{x/y}=(\frac{d}{ab})^{x/y}[/tex]

So yes, it is the same

Also:

[tex]\frac{\sqrt{10}}{\sqrt{160}}=\sqrt{\frac{10}{160}}=\sqrt{\frac{1}{16}}=\frac{1}{4}[/tex]

So you are again correct
 
Last edited:
I assume you are talking about the expression:
[tex]\frac{1}{(\frac{ab}{d})^{\frac{x}{y}}}[/tex]
By ordinary rules of arithmetic, we have:
[tex]\frac{1}{(\frac{ab}{d})^{\frac{x}{y}}}=\frac{(1)^{\frac{x}{y}}}{(\frac{ab}{d})^{\frac{x}{y}}}=[/tex]
[tex](\frac{1}{\frac{ab}{d}})^{\frac{x}{y}}=(\frac{d}{ab})^{\frac{x}{y}}[/tex]

Secondly, we have for positive, real numbers a,b:
[tex]\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}[/tex]
EDIT:
Hmm..dawillem beat me here..
 
arildno said:
I assume you are talking about the expression:
[tex]\frac{1}{(\frac{ab}{d})^{\frac{x}{y}}}[/tex]
By ordinary rules of arithmetic, we have:
[tex]\frac{1}{(\frac{ab}{d})^{\frac{x}{y}}}=\frac{(1)^{\frac{x}{y}}}{(\frac{ab}{d})^{\frac{x}{y}}}=[/tex]
[tex](\frac{1}{\frac{ab}{d}})^{\frac{x}{y}}=(\frac{d}{ab})^{\frac{x}{y}}[/tex]

Secondly, we have for positive, real numbers a,b:
[tex]\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}[/tex]
EDIT:
Hmm..dawillem beat me here..

Dear Arildno,

The bit that says by the ordinary rules of arithmetic we have...

Why did you apply the x/y to the top and bottom ?
I thought it only applies to what's inside the brackets at the bottom ?
Please can you explain this for me



Also for the last bit, on roots, if it was root minus x / root minus y what is the general answer for that ?

Thanks


Roger
 
roger said:
Why did you apply the x/y to the top and bottom ?
I thought it only applies to what's inside the brackets at the bottom ?
Please can you explain this for me

It just made it easier and 1a=1 for any value of a.
 
Response to your first question: My understanding is the expressions are equivalent, as would be
1/((ab^x/y)/d^x/y) and (d^x/y)/(ab^xy). You need to observe any hierarchy is all.

The second is numerically correct. Unless you need to express it as root 1/16 or 1/root 16 for some reason.
 

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