Solve Algebraic Equation: r > b/(sqrt2-1) = (sqrt2+1)b | Easy Explanation

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Discussion Overview

The discussion revolves around an algebraic equation derived from a problem in probability, specifically focusing on the transformation of the inequality and the equivalence between two expressions involving the variable r and a constant b. The scope includes algebraic manipulation and understanding of inequalities.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents an algebraic equation and expresses confusion about how to transform the expression r > b/(sqrt2-1) into r = (sqrt2+1)b.
  • Another participant suggests dividing both sides by (sqrt2+1) and manipulating the denominator to clarify the transformation.
  • A different participant proposes an alternative approach by multiplying both sides by (sqrt2-1) instead.
  • One participant emphasizes the importance of understanding the transformation process rather than just verifying the result, mentioning the concept of rationalizing the denominator.

Areas of Agreement / Disagreement

Participants present differing methods for manipulating the equation, indicating a lack of consensus on the preferred approach to achieve the transformation.

Contextual Notes

There are assumptions regarding the conditions under which the transformations are valid, particularly concerning the values of b and the implications of the inequalities involved.

rwinston
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Hi

I am working through the superb book "50 challenging problems in probability." I have a slight problem with some of the algebra used in solving one of the problems. The algebra to solve is shown below:

[tex] \frac{r}{r+b}\times\frac{r-1}{r+b-1}=\frac{1}{2}[/tex]

Since, for b > 0:
[tex] \frac{r}{r+b} > \frac{r-1}{r+b-1}[/tex]

[tex] \left(\frac{r}{r+b}\right)^2 > \frac{1}{2} > \left(\frac{r-1}{r+b-1}\right)^2[/tex]

Thus
[tex]\frac{r}{r+b} > \frac{1}{\sqrt{2}} > \frac{r-1}{r+b-1}[/tex]

So (this is the part I have difficulty with):

[tex]r > \frac{b}{\sqrt{2}-1} = (\sqrt{2}+1)b[/tex]

I can't see how the two sides of the = sign can be transformed into each other, or how they are equivalent. Can anyone help with this?

Cheers
 
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Divide both sides with [tex]\sqrt{2}+1[/tex] work the product in the denominator to yield one, and you'll see.
 
I would have multiplied both sides by [itex]\sqrt{2}-1[/itex]!:rolleyes:
 
But surely better than to verify, is to understand how to transform things like this in the future. It is rationalizing the denominator.
 

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