Denominator Identity Verification

Question

Here is the given information for the following proof I'm about to begin:

Suppose a, b, c, d, e and f are non-zero elements of field such that $$ \frac{a}{b} = \frac {c}{d} = \frac{e}{f}$$

I'm suppose to show the following identity is true, whenever the denominator in question is non-zero:

$$\frac{a}{b} = \frac{(a+c+e)}{(b+d+f)}$$

Here my thinking process: In order for this identity to hold, $c + e$ and $d + f$ must add up to $0$, or so I believe. Otherwise, the equality wouldn't hold true at all. Another way of thinking about this is that perhaps the letters a, c,e are multiples of one another (same with b, d, f).

You also have $af = be$ and $cf = de$. It then follows that $a(b+d+f) = ab + ad + af = ba + bc + be = b(a+ c+e)$. — joeb, Feb 28 '17 at 23:28
Basically. The part "whenever the denominator $b+d+f$ in question is nonzero" must be a hypothesis, though. It is not necessarily given that $b+d+f \neq 0$. Take $\frac{2}{2}=\frac{-1}{-1}=\frac{-1}{-1}$, for instance. — joeb, Feb 28 '17 at 23:35
Yes I was thinking of counter examples as well like yours for example. And yes, we were given that the denominators are non-zero. — John W. Smith, Feb 28 '17 at 23:36
Hint: prove first that $a/b=c/d \implies a/b=(a+c)/(b+d)$ when $b,d,b+d \ne 0,$, then apply it twice. — dxiv, Mar 01 '17 at 19:22

score 3 · Answer 1 · answered Feb 28 '17 at 23:30

3

$$\frac{a}{b}=\frac{c}{d}=\frac{e}{f}\rightarrow c=ax,d=bx,e=ay,f=by\\\frac{a+c+e}{b+d+f}=\frac{a+ax+ay}{b+bx+by}=\frac{a(1+x+y)}{b(1+x+y)}=\frac{a}{b}$$

answered Feb 28 '17 at 23:30

76david76

1,535

score 2 · Accepted Answer · answered Feb 28 '17 at 23:30

2

We have $ad=bc$, $af=be$ and $cf=de$. Now let's compute

$$a(b+d+f)-b(a+c+e)=ab+ad+af-ab-bc-be=0$$

Therefore whenever the denominators are non zero

$${a\over b}={a+c+e\over b+d+f}$$

answered Feb 28 '17 at 23:30

marwalix

16,773

Yes, this is what joeb and I were discussing up in chat. It's all clear. – John W. Smith Feb 28 '17 at 23:32

Bill Dubuque · Answer 3 · 2017-03-01T19:10:45.947

1

$$\begin{align} b\,x\, &=\, a\\ d\,x\, &=\, c\\ f\,x\, &=\, e \\ \smash[t]{\overset{\rm add}\Longrightarrow}\,\ (b\!+\!d\!+\!f)\,x\, &=\, a\!+\!c\!+\!e \end{align}$$

Said geometrically: the vectors $\,(a,b),(c,d),(e,f)\,$ have the same slope hence so does their sum, i.e. the solutions of $\, ay = bx\,$ are a submodule of $\,\Bbb Z^2,\,$ just like in the vectors space case.

edited Mar 01 '17 at 19:10

answered Feb 28 '17 at 23:46

Bill Dubuque

272,048

Denominator Identity Verification

3 Answers3