# show that if $\frac{a}{b} = \frac{c}{d}$ then $\frac{a}{b}=\frac{c}{d} = \frac{a-c}{b-d}$

$$\frac{a}{b}=\frac{c}{d} = \frac{a-c}{b-d}$$

Precalculus Mathematics in a nutshell says it's easy to verify by cross-multiplication but I'm having difficulty doing so.

(not a homework problem, any help would be appreciated)

• this makes sense if $b\ne0, d\ne0, b\ne d$ Commented Jul 7, 2019 at 14:52
• moreover $\frac{a}{b} = \frac{c}{d} = \frac{k_1 a + k_2 c}{k_1 b + k_2 d}$ Commented Jul 7, 2019 at 14:54

Substituting $$a=bt,c=dt$$ then $$\frac{a-c}{b-d}=\frac{bt-dt}{b-d}=t$$

$$\dfrac{a}{b} = \dfrac{c}{d} \Rightarrow ad = bc \Rightarrow ad - cd = bc - cd \Rightarrow (a-c)d = (b-d)c \Rightarrow \dfrac{a-c}{b-d} = \dfrac{c}{d} = \dfrac{a}{b}$$

• perfect, thank you Commented Jul 7, 2019 at 16:10
• Anybody knows if this theorem has a name ? Commented Jun 21, 2023 at 19:31

Hint: Asserting that $$\frac ab=\frac cd$$ is equivalent to $$ad=bc$$. Now, use this equality to prove that $$\frac{a-c}{b-d}=\frac ab$$.

• yeah, I spent an hour doing that last night. I will give it another stab Commented Jul 7, 2019 at 14:39
• Haven't you just rephrased the question? How does it help? Commented Jul 7, 2019 at 14:43
• @Zacky It helps a lot if the OP sees that $\frac{a-c}{b-d}=\frac ab\iff (a-c)b=(b-d)a$. Commented Jul 7, 2019 at 14:44
• Well, assuming he spent an hour already, OP didn't see that, or something is unclear. In the question he mentioned about cross mutiplication, but couldn't do it. Commented Jul 7, 2019 at 14:49
• thank you. I would have gotten from the hint. I did set up the cross multiplication but didn't equate it to a/b. thank you everyone Commented Jul 7, 2019 at 15:44

$$\dfrac a b - \dfrac {a-c}{b-d} = \dfrac {a(b-d)-b(a-c)}{b(b-d)}=\dfrac{bc-ad}{b(b-d)}=\dfrac{c-a\dfrac d b }{b-d}=\dfrac{c-a\dfrac c a}{b-d}=0$$

• this assumes $d\ne b\ne0$ Commented Jul 7, 2019 at 14:50

Simply subtract the fraction's (defining) equations, i.e.

\begin{align} b\,y &\,=\, a\\ d\,y &\,=\, c_{\phantom{|}}\\ \hline \!\!\!\!\!\!\!\!\!\!\!\!\Rightarrow\ \ (b\!-\!d)\,y &\,=\, a\!-\!c\end{align}\qquad

Said geometrically: subtraction preserves same-slope vectors (here points on the line $$\,y \,=\, (a/b)\,x)$$

Remark  This leads to a vector form of the Euclidean algorithm, and the descent step in a proof of Euclid's Lemma and the Fundamental Theorem of Arithmetic (essentially dating back to Euclid).

\begin{align} \text{If }\qquad \frac ab=\frac cd&=m\\ \text{then}\qquad\qquad a&=mb\\ c&=md\\ \therefore \frac {\lambda a+\mu c}{\lambda b+\mu d} &=\frac{\lambda mb+\mu md}{\lambda b+\mu d}\\ &=\frac {m \big(\lambda b+\mu d\big)}{\lambda b+\mu d}\\ &=m\\ \end{align}