Last Sync: 2022-04-29 08:00:47

Mathematics/Prealgebra/Mixed and improper fractions.md

@@ -0,0 +1,30 @@
+---
+tags:
+  - Mathematics
+  - Prealgebra
+  - fractions
+---
+
+# Mixed and improper fractions
+
+## Improper fractions
+
+* Proper fraction:
+  * The numerator is smaller than the denominator
+  * E.g. $\frac{2}{3}$, $-\frac{5}{10}$
+* Improper fraction
+  * The numerator is greater than the denominator
+  * E.g. $\frac{3}{2}$, $-\frac{5}{4}$
+
+## Mixed fractions
+A mixed fraction is part whole number, part fraction. For example: $5 \frac{3}{4}$.
+
+It means, in effect: $5 + \frac{3}{4}$
+
+## Converting mixed fractions into improper fractions
+Mixed fractions are hard to calculate with. We need some way to convert them to fractions. We can do this by converting them to improper fractions.
+
+With the example $4 \frac{7}{8}$, we know this means $4 + \frac{7}{8}$. We need to express the amount 4 in terms of eighths. It would be 4 lots of $\frac{8}{8}$ given that 4 is a whole number not a fractional amount. Thus the process would be:
+$$
+    \frac{8}{8} + \frac{8}{8} + \frac{8}{8} + \frac{8}{8} + \frac{7}{8}
+$$

Mathematics/Prealgebra/Multiplying_fractions.md

@@ -6,14 +6,16 @@ tags:
   - multiplication
 ---
 
+# Multiplying fractions 
 
- > 
- > To find the product of two fractions $\frac{a}{b}$ and $\frac{c}{d}$ multiply their numerators and denominators and then reduce: $$\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}$$
+ To find the product of two fractions $\frac{a}{b}$ and $\frac{c}{d}$ multiply their numerators and denominators and then reduce: 
+ 
+ $$\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}$$
 
-## Example
+###  Example
 
 $$
-\\frac{1}{3} \cdot \frac{2}{5} = \frac{1 \cdot 2}{3 \cdot 5} = \frac{2}{15}
+\frac{1}{3} \cdot \frac{2}{5} = \frac{1 \cdot 2}{3 \cdot 5} = \frac{2}{15}
 $$
 
 ## Prime factorisation in place
@@ -21,43 +23,43 @@ $$
 The example above did not require a reduction, so here is a more complex example:
 
 $$
-\\frac{14}{15} \cdot \frac{30}{140} = \frac{420}{2100} 
+\frac{14}{15} \cdot \frac{30}{140} = \frac{420}{2100} 
 $$
 
-It would be laborious to reduce such a large product using factor trees or the repeated application of divisors.  We can use a more efficient method. 
-
-This method can be applied at the point at which we conduct the multiplication rather than afterwards once we have the product. We express the the initial multiplicands as factors:
+It would be laborious to reduce such a large product using factor trees or the repeated application of divisors, as defined in [reducing fractions](./Reducing_fractions.md).  We can use a more efficient method.
+This method can be applied at the point at which we conduct the multiplication rather than afterwards once we have the product. We express the the initial multiplicands as prime factors:
 
 $$
-\\frac{14}{15} \cdot \frac{30}{140} = \frac{(2 \cdot 7) \cdot (2 \cdot 3 \cdot 5) }{(3 \cdot 5) \cdot (2 \cdot  2 \cdot 7 \cdot 5)} 
+\frac{14}{15} \cdot \frac{30}{140} = \frac{(2 \cdot 7) \cdot (2 \cdot 3 \cdot 5) }{(3 \cdot 5) \cdot (2 \cdot  2 \cdot 7 \cdot 5)} 
 $$
 
 We now have the product in factorised form before we have applied the multiplication so we can go ahead and cancel: 
 
 $$
-\\frac{\cancel{2}, \cancel{7}, \cancel{2}, \cancel{3}, \cancel{5}}{\cancel{3}, \cancel{5}, \cancel{2}, \cancel{2}, \cancel{7}, 5} = \frac{1}{5}
+\frac{\cancel{2}, \cancel{7}, \cancel{2}, \cancel{3}, \cancel{5}}{\cancel{3}, \cancel{5}, \cancel{2}, \cancel{2}, \cancel{7}, 5} = \frac{1}{5}
 $$
 
 **Note that in the above case, there was only a single 5 left as a denominator and no value left as a numerator. This is equivalent to there just being "one five" so we write $\frac{1}{5}$**
 
-## Example with negative fractions containing variables
+## Example of multiplying fractions with negative fractions containing variables
 
-*Calculate: $$ - \frac{6x}{55y} \cdot - \frac{110y^2}{105x^2} $$*
+Calculate: 
+$$- \frac{6x}{55y} \cdot - \frac{110y^2}{105x^2}$$
 
 First multiply in place: 
 $$ 
-\\frac{(3 \cdot 2 \cdot x) \cdot (5  \cdot 2 \cdot 11 \cdot y \cdot y)}{(5 \cdot 11 \cdot y) \cdot (7 \cdot 5 \cdot 3 \cdot x \cdot x)}
+\frac{(3 \cdot 2 \cdot x) \cdot (5  \cdot 2 \cdot 11 \cdot y \cdot y)}{(5 \cdot 11 \cdot y) \cdot (7 \cdot 5 \cdot 3 \cdot x \cdot x)}
 $$
 
 Then cancel: 
 
 $$ 
-\\frac{(\cancel{3} \cdot 2 \cdot \cancel{x}) \cdot (\cancel{5}  \cdot 2 \cdot \cancel{11} \cdot \cancel{y} \cdot y)}{(\cancel{5} \cdot \cancel{11} \cdot \cancel{y}) \cdot (7 \cdot 5 \cdot \cancel{3} \cdot \cancel{x} \cdot x)} = 
-\\frac{2  \cdot 2 \cdot y}{7 \cdot 5 \cdot x}
+\frac{(\cancel{3} \cdot 2 \cdot \cancel{x}) \cdot (\cancel{5}  \cdot 2 \cdot \cancel{11} \cdot \cancel{y} \cdot y)}{(\cancel{5} \cdot \cancel{11} \cdot \cancel{y}) \cdot (7 \cdot 5 \cdot \cancel{3} \cdot \cancel{x} \cdot x)} = 
+\frac{2  \cdot 2 \cdot y}{7 \cdot 5 \cdot x}
 $$
 
 Then reduce: 
 
 $$
-\\frac{2  \cdot 2 \cdot y}{7 \cdot 5 \cdot x} = \frac{4y}{35x} 
+\frac{2  \cdot 2 \cdot y}{7 \cdot 5 \cdot x} = \frac{4y}{35x} 
 $$