What identity is this? (Division to multiplication )

AI Thread Summary
The discussion centers on the identities related to the transformation of division into multiplication. It clarifies that there is no direct identity that converts a/b to ab or A^2/B^2 to (A^2)(B^2). Instead, it highlights that dividing by b can be expressed as multiplying by its reciprocal, 1/b. Additionally, it mentions the Newton-Raphson method for approximating division through iterative multiplication and addition. Ultimately, the inquiry confirms that such transformations are not standard identities.
musiclover55
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This isn't a homework help issue, I just want to know what identity(?) this is.

a/b to ab

or A^2/B^2 to (A^2)(B^2)
 
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musiclover55 said:
This isn't a homework help issue, I just want to know what identity(?) this is.

a/b to ab

or A^2/B^2 to (A^2)(B^2)
There is no identity that converts a/b to ab, nor is there one to convert A2/B2 to A2B2.

What you might be thinking of is this one:

$$ \frac{a}{b} = a \cdot \frac{1}{b}$$

In other words, dividing by a number b is the same as multiplying by the reciprocal of b (1/b).
 
Mark44 said:
What you might be thinking of is this one:

$$ \frac{a}{b} = a \cdot \frac{1}{b}$$

Or maybe this one:$$
\frac a {\frac 1 b} = ab$$
 
Although this probably isn't what you're looking for, there is a way to implement division or more specifically 1/b by iterating multiplications and additions via Newton Raphson method, but this seems unlikely to be what you're looking for.

For 1/b the iternation is:

xi+1 = xi (2 - b xi)

This converges quadratically. For integers, usually a table of bytes is used to get the initial approximation.
 
Yea, thanks everyone. I thought that was possible, but obviously it's not lol.
 
Yet another possibility: A/B= AB-1
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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