- #201

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There is an intimate relation between two quantities such that for all positive values between zero and infinity of the independent variable, two distinct graphs can be shown that the product of ab=a-b.

[tex] a=\frac{b}{1-b}[/tex]

[tex] b=\frac{a}{1+a}[/tex]

[tex] 1-b=\frac{b}{a}[/tex]

[tex] 1+a=\frac{a}{b}[/tex]

[tex](1-b)(1+a)=1[/tex]

[tex] ab=a-b[/tex]

The graph of abscissa b versus ordinate a is a discontinuous curve at 1, with 0 to infinity and from negative infinity approaching -1.

The graph of abscissa a versus ordinate b is a curve from 0 and approaching the value 1 at infinity.

[tex] a=\frac{b}{1-b}[/tex]

[tex] b=\frac{a}{1+a}[/tex]

[tex] 1-b=\frac{b}{a}[/tex]

[tex] 1+a=\frac{a}{b}[/tex]

[tex](1-b)(1+a)=1[/tex]

[tex] ab=a-b[/tex]

The graph of abscissa b versus ordinate a is a discontinuous curve at 1, with 0 to infinity and from negative infinity approaching -1.

The graph of abscissa a versus ordinate b is a curve from 0 and approaching the value 1 at infinity.

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