Hölder's Theorem & Manipulating 1/p + 1/q = 1

[zxh]

Homework Statement

How do you get from 1/p +1/q = 1 to (p-1)(q-1) = 1? (http://de.wikipedia.org/wiki/Konjugation_%28Reelle_Zahlen%29" )

Homework Equations

Is log(xy) = log(x) + log(y) in any way related to this, i.e the rewritten subject title?

The Attempt at a Solution

Manipulate[
ContourPlot[x + y == x*y, {x, -m, m}, {y, -m, m},
PlotRange -> Automatic], {m, -100, 100}]

TIA!

 

[disregardthat]

Multiply out and compare. Logarithms have nothing to do with the equation you mentioned, but you might be using the logarithm to prove some other aspect of Hölders inequality.

 

[ehild]

How do you get from 1/p +1/q = 1 to (p-1)(q-1) = 1? (http://de.wikipedia.org/wiki/Konjugation_%28Reelle_Zahlen%29" )

You can get the second equation from the first one by multiplying both sides of the first equation by pq, then subtracting 1, and factoring. Try, it is fun.

ehild

 

[zxh]

Alright, makes me look tired, but i thought the similarity with the logarithm was too salient not to connect the two.