Where can I find resources regarding separating variables from fractions.

[hemmed]

Hi there,

I am currently working through Algebra Demystified and have come across some issues understanding the following:

At times, especially in calculus, students need to separate a variable from the rest of the fraction. This involves writing the fraction as a product of two fractions, or of one fraction and a whole number, or of one fraction and a variable.

Specifically, the book provides the following rules

. . . . .$\displaystyle \large{ \dfrac{a}{b}\, =\, a\, \cdot\, \dfrac{1}{b} }$

. . . . .$\displaystyle \large{\dfrac{a}{b}\, =\, \dfrac{1}{b}\, \cdot\, a}$

. . . . .$\displaystyle \large{\dfrac{ab}{c}\, =\, a\, \cdot\, \dfrac{b}{c}\, =\, \dfrac{a}{c}\, \cdot\, b}$

. . . . .$\displaystyle \large{\dfrac{a}{bc}\, =\, a\, \cdot\, \dfrac{1}{bc}\, =\, \dfrac{a}{b}\, \cdot\, \dfrac{1}{c}}$

I am really struggling to understand these - I can follow it from memory but I'd really like to be able to grasp the theory of this topic more. On a basic level, this makes sense to me:

. . . . .$\displaystyle \large{\dfrac{x}{3}\, =\, \dfrac{1\, \cdot\, x}{3\, \cdot\, 1}\, =\, \dfrac{1}{3}\, \cdot\, \dfrac{x}{1}\, =\, \dfrac{1}{3}\,x}$

. . . . .$\displaystyle \large{\dfrac{a}{b}\, =\, \dfrac{1}{b}\, \cdot\, a}$

But I am struggling with more complex fractions. Are there any good guides online which can explain in more detail how this works?

Thanks in advance.

 

[Steven G]

Hi there,

I am currently working through Algebra Demystified and have come across some issues understanding the following:

At times, especially in calculus, students need to separate a variable from the rest of the fraction. This involves writing the fraction as a product of two fractions, or of one fraction and a whole number, or of one fraction and a variable.

Specifically, the book provides the following rules

. . . . .$\displaystyle \large{ \dfrac{a}{b}\, =\, a\, \cdot\, \dfrac{1}{b} }$

. . . . .$\displaystyle \large{\dfrac{a}{b}\, =\, \dfrac{1}{b}\, \cdot\, a}$

. . . . .$\displaystyle \large{\dfrac{ab}{c}\, =\, a\, \cdot\, \dfrac{b}{c}\, =\, \dfrac{a}{c}\, \cdot\, b}$

. . . . .$\displaystyle \large{\dfrac{a}{bc}\, =\, a\, \cdot\, \dfrac{1}{bc}\, =\, \dfrac{a}{b}\, \cdot\, \dfrac{1}{c}}$

I am really struggling to understand these - I can follow it from memory but I'd really like to be able to grasp the theory of this topic more. On a basic level, this makes sense to me:

. . . . .$\displaystyle \large{\dfrac{x}{3}\, =\, \dfrac{1\, \cdot\, x}{3\, \cdot\, 1}\, =\, \dfrac{1}{3}\, \cdot\, \dfrac{x}{1}\, =\, \dfrac{1}{3}\,x}$

. . . . .$\displaystyle \large{\dfrac{a}{b}\, =\, \dfrac{1}{b}\, \cdot\, a}$

But I am struggling with more complex fractions. Are there any good guides online which can explain in more detail how this works?

Thanks in advance.

Any number (even a fraction) is equal to itself divided by one. When multiplying two fractions just multiply the numerator and put the result in the numerator and multiply the denominators and put the result in the denominator.

$\displaystyle So \ \ why \ \ is \ \ a*\dfrac{b}{c} = \dfrac{ab}{c} \ \ is \ \ your \ \ question?$

To see this we need to use the associative law and order of operations which says we do multiplication and division from left to right. Please let ** represent the division symbol

$\displaystyle (a)(\dfrac{b}{c}) = (a)(b**c)= (ab)**c =\dfrac{ab}{c}$

 

[Otis]

Specifically, the book provides the following rules

. . . . .$\displaystyle \large{ \dfrac{a}{b}\, =\, a\, \cdot\, \dfrac{1}{b} }$

. . . . .$\displaystyle \large{\dfrac{a}{b}\, =\, \dfrac{1}{b}\, \cdot\, a}$

. . . . .$\displaystyle \large{\dfrac{ab}{c}\, =\, a\, \cdot\, \dfrac{b}{c}\, =\, \dfrac{a}{c}\, \cdot\, b}$

. . . . .$\displaystyle \large{\dfrac{a}{bc}\, =\, a\, \cdot\, \dfrac{1}{bc}\, =\, \dfrac{a}{b}\, \cdot\, \dfrac{1}{c}}$

The first two factorizations show the Commutative Property of Multiplication. That is, we can multiply numbers in any order and the result is the same.

The third shows that any factor within a numerator can be factored out of the ratio. Think of factoring ratios as "undoing" multiplication. In other words, if you start with two ratios that are being multiplied together:

a/c * b/1

then their product is:

(ab)/c

because the rule for multiplying two ratios is (numerator times numerator) over (denominator times denominator).

Now, because these two expressions in red are equal, if you started out with the form (ab)/c instead, you can always return to the factored form a/c*b/1, if that's handy.

The fourth shows that any factor within a denominator can also be factored out, using 1 as the numerator. There's another possibility: undo all of the multiplications:

a/(bc) = a/1 * 1/b * 1/c

From this decomposition, you could multiply the first two factors, to arrive at:

a/b * 1/c

Or, you could multiply the last two factors, to arrive at:

a/1 * 1/(bc)

I can follow it from memory but I'd really like to be able to grasp the theory of this topic more.

Good for you! Learning math is more than memorizing a bunch of rules.

I think you just need more exposure. Once you've practiced enough, factoring algebraic ratios will come more naturally (without having to memorize a form for each situation) because they all come apart in basically the same ways.

 

[hemmed]

The first two factorizations show the Commutative Property of Multiplication. That is, we can multiply numbers in any order and the result is the same.

The third shows that any factor within a numerator can be factored out of the ratio. Think of factoring ratios as "undoing" multiplication. In other words, if you start with two ratios that are being multiplied together:

a/c * b/1

then their product is:

(ab)/c

because the rule for multiplying two ratios is (numerator times numerator) over (denominator times denominator).

Now, because these two expressions in red are equal, if you started out with the form (ab)/c instead, you can always return to the factored form a/c*b/1, if that's handy.

The fourth shows that any factor within a denominator can also be factored out, using 1 as the numerator. There's another possibility: undo all of the multiplications:

a/(bc) = a/1 * 1/b * 1/c

From this decomposition, you could multiply the first two factors, to arrive at:

a/b * 1/c

Or, you could multiply the last two factors, to arrive at:

a/1 * 1/(bc)

This is such a perfect explanation, I understand it loads better now!