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On page 165 of Chapter 13, how was the equality made from line 1 to line 2?

https://archive.org/details/NumberTheory_862/page/n173

Namely, how $$ \prod_{j=1}^{\infty} \frac{1}{1-q^j} = \prod_{j=1}^{\infty} \frac{1}{(1-q^{2j-1})(1-q^{2j})}$$

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Simply pair-off factors and use the fact that multiplication is commutative: \begin{align*} \prod_{j=1}^{\infty}\frac{1}{(1-q^{j})} & = \frac{1}{1-q}\frac{1}{1-q^2}\frac{1}{1-q^3}\frac{1}{1-q^4}\cdots\\ & = \left(\frac{1}{1-q}\frac{1}{1-q^2}\right)\left(\frac{1}{1-q^3}\frac{1}{1-q^4}\right)\cdots\\ & = \left(\frac{1}{1-q^{2}}\frac{1}{1-q}\right)\left(\frac{1}{1-q^4}\frac{1}{1-q^{3}}\right)\cdots\\ & = \prod_{j=1}^{\infty}\frac{1}{(1-q^{2j})(1-q^{2j-1})}. \end{align*}

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  • $\begingroup$ Oh, I see. Thank you. $\endgroup$ – zodross Dec 17 '18 at 0:32

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