Let \((a, c) \in \hspace{0.05em} \mathbb{R}^2 \) be two real numbers, and \((b, d) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^2 \) two non-zero real numbers.
$$ \forall (a, c) \in \hspace{0.05em} \mathbb{R}^2, \enspace (b, d) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^2, $$
$$ \frac{a}{b} = \frac{c}{d} \Longleftrightarrow ad = bc $$
Ratio between respectives numerators and denominators
$$ \forall (a, b, c,d) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^4, \ \Bigl \{ (a+b),(c+d) \Bigr \} \ \neq 0,$$
$$ \frac{a}{b} = \frac{c}{d} \Longleftrightarrow \frac{a + b}{a} = \frac{c+d}{c} \Longleftrightarrow \frac{a}{a + b} = \frac{c}{c + d} $$
$$ \frac{a}{b} = \frac{c}{d} \Longleftrightarrow \frac{a+b}{b} = \frac{c+d}{d} \Longleftrightarrow \frac{b}{a + b} = \frac{d}{c + d} $$
The same ratios are possibles replacing all \( (+) \) by \( (-) \).
Ratio between respectives sums an differences
$$ \forall (a, c) \in \hspace{0.05em} \mathbb{R}^2, \enspace (b, d) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^2, \Bigl \{ (a-b), (c-d) \Bigr \} \ \neq 0, $$
$$ \frac{a}{b} = \frac{c}{d} \Longleftrightarrow \frac{a+b}{a-b} = \frac{c+d}{c-d} $$
Addition of numerators and denominators
$$ \forall (a, c) \in \hspace{0.05em} \mathbb{R}^2, \enspace (b, d) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^2, \enspace \ \Bigl \{ (b+d) \Bigr \} \ \neq 0, $$
$$ \frac{a}{b} = \frac{c}{d} = \frac{a+c}{b+d}$$
The same relation is possible replacing \( (+) \) by \( (-) \).
On the whole, with a serie of \(n \) numerators and \(m \) denominators:
$$ \forall (a, c, e ...) \in \hspace{0.05em} \mathbb{R}^n, \enspace (b, d, f...) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^m, \enspace \ \Bigl \{ (b\pm d \pm f ...) \Bigr \} \ \neq 0, $$
$$ \frac{a}{b} = \frac{c}{d} = \frac{e}{f} = \ ... \ = \frac{a\pm c \pm e \pm \ ...}{b\pm d \pm f \pm \ ...}$$
Recap table of the properties formulas of fractions
Let \((a, c) \in \hspace{0.05em} \mathbb{R}^2 \) be two real numbers, and \( (b, d) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^2 \) two non-zero real numbers.
And let \((H)\) be the following hypothesis:
$$ \forall (a, c) \in \hspace{0.05em} \mathbb{R}^2, \enspace (b, d) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^2, \hspace{1em} \frac{a}{b} = \frac{c}{d} \qquad (H) $$
Starting from the hypothesis \((H)\):
$$ \forall (a, c) \in \hspace{0.05em} \mathbb{R}^2, \enspace (b, d) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^2, \hspace{1em} \frac{a}{b} = \frac{c}{d} \qquad (H) $$
We multiply both members by the same number \( bd \):
$$ \frac{abd}{b} = \frac{cbd}{d} $$
$$ ad = cb $$
And finally,
$$ \forall (a, c) \in \hspace{0.05em} \mathbb{R}^2, \enspace (b, d) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^2, $$
$$ \frac{a}{b} = \frac{c}{d} \Longleftrightarrow ad = bc $$
Starting from the hypothesis \((H)\):
$$ \forall (a, c) \in \hspace{0.05em} \mathbb{R}^2, \enspace (b, d) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^2, \hspace{1em} \frac{a}{b} = \frac{c}{d} \qquad (H) $$
With the result of the cross product we do have:
$$ \frac{a}{b} = \frac{c}{d} \Longleftrightarrow ad = bc $$
Let us add \( ac \) on both sides of the equation:
$$ ad + ac = bc + ac $$
We factorize it:
$$ a(d+c) = (a+b)c $$
$$ \frac{a}{a + b} = \frac{c}{c + d} \qquad(1) $$
Moreover, rearranging \((1)\) by applying the inverse on both sides:
$$ \frac{a + b}{a} = \frac{c+d}{c} \qquad(2) $$
Now, both expressions \((1)\) and \((2)\) had to support a new existing condition to be compliant:
$$\Bigl \{ a, c, (a+b),(c+d) \Bigl \} \ \neq 0 $$
As a result we do have,
$$ \forall (a, b, c,d) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^4, \ \Bigl \{ (a+b),(c+d) \Bigr \} \ \neq 0,$$
$$ \frac{a}{b} = \frac{c}{d} \Longleftrightarrow \frac{a + b}{a} = \frac{c+d}{c} \Longleftrightarrow \frac{a}{a + b} = \frac{c}{c + d} $$
Consequently, reproducing the same exact reasoning adding now the term \( bd \), we obtain the following result:
$$ \forall (a, b, c,d) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^4, \ \Bigl \{ (a+b),(c+d) \Bigr \} \ \neq 0,$$
$$ \frac{a}{b} = \frac{c}{d} \Longleftrightarrow \frac{a+b}{b} = \frac{c+d}{d} \Longleftrightarrow \frac{b}{a + b} = \frac{d}{c + d} $$
It is possible to find out the same expressions replacing all \( (+) \) by \( (-) \), not by adding terms but by removing it.
$$ \forall (a, b, c,d) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^4, \ \Bigl \{ (a\textcolor{#A65757}{-}b),(c\textcolor{#A65757}{-}d) \Bigr \} \ \neq 0,$$
$$ \frac{a}{b} = \frac{c}{d} \Longleftrightarrow \frac{a \textcolor{#A65757}{-} b}{a} = \frac{c\textcolor{#A65757}{-}d}{c} \Longleftrightarrow \frac{a}{a \textcolor{#A65757}{-} b} = \frac{c}{c \textcolor{#A65757}{-} d} $$
$$ \forall (a, b, c,d) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^4, \ \Bigl \{ (a\textcolor{#A65757}{-}b),(c\textcolor{#A65757}{-}d) \Bigr \} \ \neq 0,$$
$$ \frac{a}{b} = \frac{c}{d} \Longleftrightarrow \frac{a\textcolor{#A65757}{-}b}{b} = \frac{c\textcolor{#A65757}{-}d}{d} \Longleftrightarrow \frac{b}{a \textcolor{#A65757}{-} b} = \frac{d}{c \textcolor{#A65757}{-} d} $$
Starting from the hypothesis \((H)\):
$$ \forall (a, c) \in \hspace{0.05em} \mathbb{R}^2, \enspace (b, d) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^2, \hspace{1em} \frac{a}{b} = \frac{c}{d} \qquad (H) $$
With the result of the cross product we do have:
$$ \frac{a}{b} = \frac{c}{d} \Longleftrightarrow ad = bc $$
Let us add both terms \( ac \) and \( (-bd) \) on both sides of the equation:
$$ ad + ac - bd = bc + ac -bd $$
Then, as our hypothesis is that \( ad = bc \), we can add \( (-bc) \) on one side and add \( (-ad) \) on another.
$$ ad + ac - bd - bc = bc + ac -bd - ad $$
We factorize by \( a\) and also by \( b\) to obtain:
$$ a(c+d) - b(c+d) = a(c-d) + b(c-d) $$
$$ (a-b)(c+d) = (a+b)(c-d) $$
$$ \frac{a+b}{a-b} = \frac{c+d}{c-d} \qquad(3) $$
But, the expression \((3)\) has to support a new existing condition to be compliant:
$$\Bigl \{ (a-b), (c-d) \Bigl \} \ \neq 0 $$
On the whole, depending of which factor is on the denominator, we had to add the good existing condition to it to be non-zero.
Finally we do have,
$$ \forall (a, c) \in \hspace{0.05em} \mathbb{R}^2, \enspace (b, d) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^2,, \Bigl \{ (a-b), (c-d) \Bigr \} \ \neq 0, $$
$$ \frac{a}{b} = \frac{c}{d} \Longleftrightarrow \frac{a+b}{a-b} = \frac{c+d}{c-d} $$
Starting from the hypothesis \((H)\):
$$ \forall (a, c) \in \hspace{0.05em} \mathbb{R}^2, \enspace (b, d) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^2, \hspace{1em} \frac{a}{b} = \frac{c}{d} \qquad (H) $$
With the result of the cross product we do have:
$$ \frac{a}{b} = \frac{c}{d} \Longleftrightarrow ad = bc $$
Let us add \( cd \) on both sides of the equation:
$$ ad + cd = bc + cd $$
$$ d(a + c) = c(b + d) $$
$$ \frac{a + c}{b + d} = \frac{c}{d} $$
And finally we do have,
$$ \forall (a, c) \in \hspace{0.05em} \mathbb{R}^2, \enspace (b, d) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^2, \enspace \ \Bigl \{ (b+d) \Bigr \} \ \neq 0, $$
$$ \frac{a}{b} = \frac{c}{d} = \frac{a+c}{b+d}$$
The same relation is possible replacing all \( (+) \) by \( (-) \), not by adding \( cd \) but by removing it.
On the whole, with a serie of \(n \) numerators and \(m \) denominators:
$$ \forall (a, c, e ...) \in \hspace{0.05em} \mathbb{R}^n, \enspace (b, d, f...) \in \hspace{0.05em} \bigl[\mathbb{R}^* \bigr]^m, \enspace \ \Bigl \{ (b\pm d \pm f ...) \Bigr \} \ \neq 0, $$
$$ \frac{a}{b} = \frac{c}{d} = \frac{e}{f} = \ ... \ = \frac{a\pm c \pm e \pm \ ...}{b\pm d \pm f \pm \ ...}$$
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