Which of the statements are false?

Question

I have this statement:

Let $a, b, c, d \in \mathbb{R} - $ {$0$}, with $\quad acd> 0$.

If $– 1 < \frac{a}{b} < \frac{b}{c} < \frac{c}{d} < \frac{d}{a} < 1,$

Which of the following alternative are false?

A) $\quad a < cd$

B) $\quad c < ad$

C) $\quad d < ac$

D) $\quad ab < cd$

E) $\quad ad < bc$

I'm really stuck, I've tried several things and nothing has worked for me. Some hint or guide would be useful

real numbers other different of zero... I'm not sure what that means. Are you saying they're all nonzero? — lurker, Jun 14 '19 at 03:12
What is "different of zero"? Do you mean "different from zero"? — user10354138, Jun 14 '19 at 03:21

user10354138 · Answer 1 · 2019-06-14T03:31:02.693

2

Assuming by "different of zero" you mean $abcd\neq 0$.

We have $a/c=(a/b)(b/c)$ so $-1<a/c<1$, and we are also given $-1<c/d<1$, $-1<d/a<1$. Multiplying gives the contradiction $1=(acd)/(acd)<1$ hence there are no $a,b,c,d$ that satisfy the given inequality. Hence all options are vacuously true.

edited Jun 14 '19 at 03:31

answered Jun 14 '19 at 03:25

user10354138

33,239

I mean that $a,b,c,d, \in \mathbb{R} -$ {$0$} – ESCM Jun 14 '19 at 03:27
And the answer must be $E)$ – ESCM Jun 14 '19 at 03:29
No, all of (A) - (E) are true, because there are no counterexamples. – user10354138 Jun 14 '19 at 03:30
@EduardoS. in your question, what do you mean when you use "$<$"? $<$ or $\leq$? – dcolazin Jun 14 '19 at 03:30
Some explanation of why $E$ is false is: 'Because $ad$ is positive and $bc$ negative'. And i don't know why. – ESCM Jun 14 '19 at 03:32
@dcolazin i write the question in a correct form – ESCM Jun 14 '19 at 03:32
So give me a witness to the fact that $bc$ is negative, because it is also vacuously true that $bc$ is positive. – user10354138 Jun 14 '19 at 03:36
I do not know. That is written in the answers. There's a development in Spanish, but I do not know how to translate it since I'm not native English, and that's why I'm looking for help here. However, here is the development in Spanish: https://i.imgur.com/vuQgd1r.png – ESCM Jun 14 '19 at 03:40
Thanks for your help – ESCM Jun 14 '19 at 03:40

score 0 · Answer 2 · answered Jun 14 '19 at 05:52

Check where $0$ can fit ($5$ options): $$– 1 \overbrace{<}^{<0} \frac{a}{b} \overbrace{<}^{<0} \frac{b}{c} \overbrace{<}^{<0} \frac{c}{d} \overbrace{<}^{<0} \frac{d}{a} \overbrace{<}^{<0} 1,$$ For each case, by checking signs of the numbers $a,b,c,d$, one can find only the central sign works. For example: $$– 1 <0< \frac{a}{b} < \frac{b}{c} < \frac{c}{d} < \frac{d}{a} < 1 \Rightarrow \\ a<b<c<d<a \ ?$$ Similarly, you can reject the rest three cases by contradiction.

So we have: $$– 1 < \frac{a}{b} < \frac{b}{c} < 0<\frac{c}{d} < \frac{d}{a} < 1 \Rightarrow \\ 1) \ (a,b,c,d)=(+,-,+,+); \ \ 2) \ (a,b,c,d)=(-,+,-,-) \Rightarrow \\ 1) \ E) \ ad<bc \iff (+)(+)<(-)(+) \ ? \ 2) \ E) \ ad<bc \iff (-)(-)<(+)(-) \ ?$$ I leave other options to check to you.

Which of the statements are false?

2 Answers2