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So I'm trying to prove something, I came across a sub-question that lead me to another question I thought of. Is the following legal?

$$\prod_{k=1}^{n} \left(1-\frac{1}{2k}\right)/ \prod_{k=1}^{n} \left(1-\frac{1}{2k+1}\right) = \prod_{k=1}^{n} \left(\frac{1-\frac{1}{2k}}{1-\frac{1}{2k+1}}\right) = \prod_{k=1}^{n} \left(1-\frac{1}{4k^2}\right)$$ (Which is the answer at the back)

EDIT: Sorry guys, I mean to say -

If this is true?

$$\prod_{k=1}^{n} \left(a_{k}\right)/ \prod_{k=1}^{n} \left(b_{k}\right) = \prod_{k=1}^{n} \left(\frac{a_{k}}{b_{k}}\right) $$

If so, why? It works out for the proof I am doing, but I cannot see mathematically why? Anyone care to explain? Thank you.

Sorry the the small code, I don't know why it turned out this way.

Bobby
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    It's okay, the reason is commutativity and associativity, it's like $(a\cdot b)/(c\cdot d) = \frac{a}{c}\cdot \frac{b}{d}$, only a few more factors. – Daniel Fischer Aug 23 '13 at 15:42

2 Answers2

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Did you try to expand the notation to understand?

$$\frac{\prod_{k=1}^{n} (1-\frac{1}{2k})}{\prod_{k=1}^{n}(1 - \frac{1}{2k+1})} = \frac{(1-\frac{1}{2})(1-\frac{1}{4}) \times \ldots \times (1-\frac{1}{2n})}{(1-\frac{1}{3})(1-\frac{1}{5}) \times \ldots \times (1-\frac{1}{2n+1})}$$, and you must know that multiplying fractions consists in multiplying numerators and denominators, so regroup each term and split into $n$ different fractions!

MJD
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zarathustra
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This is just a special case of the rule for multiplying fractions: Multiply the numerators and multiply the denominators. The right side of your equation is a product of fractions, and the left side is the result of applying the rule.

Edit: While I was writing this, the question was being enlarged to add another term to the equation. What I called the "right side" has become the "middle side".

Andreas Blass
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