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

Question

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

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)

moreover $\frac{a}{b} = \frac{c}{d} = \frac{k_1 a + k_2 c}{k_1 b + k_2 d}$ — Eugene Sirkiza, Jul 07 '19 at 14:54

score 4 · Answer 1 · answered Jul 07 '19 at 14:40

4

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

answered Jul 07 '19 at 14:40

Dr. Sonnhard Graubner

95,283

score 3 · Accepted Answer · answered Jul 07 '19 at 14:46

3

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

answered Jul 07 '19 at 14:46

P.G

662

1

perfect, thank you – naftalimich Jul 07 '19 at 16:10
Anybody knows if this theorem has a name ? – jozinho22 Jun 21 '23 at 19:31

score 1 · Answer 3 · answered Jul 07 '19 at 14:38

1

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

answered Jul 07 '19 at 14:38

José Carlos Santos

427,504

yeah, I spent an hour doing that last night. I will give it another stab – naftalimich Jul 07 '19 at 14:39
Haven't you just rephrased the question? How does it help? – Zacky Jul 07 '19 at 14:43
@Zacky It helps a lot if the OP sees that $\frac{a-c}{b-d}=\frac ab\iff (a-c)b=(b-d)a$. – José Carlos Santos Jul 07 '19 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. – Zacky Jul 07 '19 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 – naftalimich Jul 07 '19 at 15:44
I suggest that you mark one of the answers as the accepted one. – José Carlos Santos Jul 07 '19 at 15:57
i will. I'm studying them now – naftalimich Jul 07 '19 at 16:08

score 1 · Answer 4 · answered Jul 07 '19 at 14:49

1

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

answered Jul 07 '19 at 14:49

J. W. Tanner

60,406

this assumes $d\ne b\ne0$ – J. W. Tanner Jul 07 '19 at 14:50

Bill Dubuque · Answer 5 · 2019-07-07T15:42:39.567

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).

score 0 · Answer 6 · answered Jul 07 '19 at 18:00

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

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

6 Answers6