[tex]\frac{a}{b} = \frac{c}{d}[/tex]
A) [tex] \frac{a+b}{b} = \frac{a}{b} + \frac{b}{b} =\frac{a}{b} + 1=\frac{c}{d} + 1 =\frac{c}{d} + \frac{d}{d} =\frac{c+d}{d} [/tex]
B) [tex] \frac{a}{a+b} = \frac{b*a}{b(a+b)}= \frac{ \frac{a}{b}}{ \frac{a+b}{b}}=\frac{ \frac{a}{b}}{ \frac{a}{b}+1}=\frac{ \frac{c}{d}}{ \frac{c}{d}+1}= \frac{ \frac{c}{d}}{ \frac{c+d}{d}}= \frac{c*d}{d(c+d)}= \frac{c}{c+d}[/tex]
C) [tex] \frac{a-b}{b} = \frac{a}{b} - \frac{b}{b} =\frac{a}{b} - 1=\frac{c}{d} - 1 =\frac{c}{d} - \frac{d}{d} =\frac{c-d}{d} [/tex]
pentru a ≥ b si c≥ d
D) [tex] \frac{a}{b-a} = \frac{b*a}{b(b-a)}= \frac{ \frac{a}{b}}{ \frac{b-a}{b}}=\frac{ \frac{a}{b}}{1- \frac{a}{b}}=\frac{ \frac{c}{d}}{1- \frac{c}{d}}= \frac{ \frac{c}{d}}{ \frac{d-c}{d}}= \frac{c*d}{d(d-c)}= \frac{c}{d-c}[/tex]
pentru a < b si c < d
E)[tex] \frac{a^n}{b^n} = (\frac{a}{b})^n= (\frac{c}{d})^n= \frac{c^n}{d^n}[/tex]
pentru orice n ∈ N
F) a+d=b+c
[tex]\frac{a}{b} = \frac{c}{d}[/tex]=>ad=bc
[tex] \frac{a+n}{b+n} = \frac{a+n}{b+n}* \frac{d+n}{d+n}=\frac{ad+an+dn+n^2}{(b+n)(d+n)}=\frac{ad+(a+d)n+n^2}{(b+n)(d+n)}=[/tex]
[tex]=\frac{bc+(b+c)n+n^2}{(b+n)(d+n)}=\frac{bc+bn+cn+n^2}{(b+n)(d+n)}=\frac{(b+n)(c+n)}{(b+n)(d+n)}=\frac{c+n}{d+n}[/tex]
pentru orice n ∈ N de
