0
$\begingroup$

Question: Given that $x=\frac{a}{\frac{a}{x+b}+c+b}+c$ for all real $x\neq -b $, prove that $a\neq 0$ and that $c=-b$.

Proving that $a\neq 0$ probably involves a proof by contradiction, but I am not sure exactly what this involves. I also think this can be shown without proving that $c=-b$, but $a\neq 0$ follows directly if we can prove that first.

To prove that $c=-b$, the first equation can be manipulated to obtain the result $a(b+c)=(x-c)(x+b)(b+c)$. We can safely divide both sides by $b+c$ as if $b+c=0$, $b=-c$ is satisfied anyway. This gets that $a=(x-c)(x+b)$. If we can prove that $a\neq 0$, then it follows that $x-c\neq 0$, and so $x\neq c$. This, because the only original restraint on $x$ was that $x\neq -b$, suggests that $c=-b$, but I am not sure this logic is rigorous or correct.

If someone can prove that $a\neq 0$ and the logic at the end is valid, then I think the problem is solved, unless I made a mistake elsewhere.

Also, this problem is only designed for strong year 10 students, so the solution is unlikely to involve much more than simple proof methods and basic algebraic manipulation.

$\endgroup$
1
$\begingroup$

I shall assume that you want the statement to hold true for all $x$. Suppose on the contrary that $a=0$, then we have $x=c$, but we can let $x$ takes two different values, $x_1 \ne x_2$ and we obtain a contradiction since we will be having $x_1 =c=c_2$. Hence $a \ne 0$.

$$x-c=\frac{a}{\frac{a}{x+b}+b+c}$$

$$x-c=\frac{a(x+b)}{a+(b+c)(x+b)}$$

$$(x-c)[a+(b+c)(x+b)]=a(x+b)$$

Suppose $c \ne -b$, and the only condition on $x$ is $x \ne -b$, then we can let $x=c$.

$$0=a(b+c)$$

Since $a \ne 0$, we have $b+c=0$.

$\endgroup$
  • $\begingroup$ I can't quite follow your explanation for why $a\neq 0$, but if it can be clarified, then the second part follows well. Thanks. $\endgroup$ – Jonathan Zhou May 4 '18 at 10:00
  • $\begingroup$ If $a=0$, then you have for all $x$, $x=c$, that is every number that is not equal to $-b$ is equal to $c$, which is a contradiction right. For example if $-b=0$ and $c=1$, you can let $x=2$ and you get $1=2$ which is a contradiction. $\endgroup$ – Siong Thye Goh May 4 '18 at 16:13
  • $\begingroup$ Thanks, makes sense now. Now it seems we can just assume that $c\neq -b$ so we can let $x=c$, which makes $0=\frac{a}{\frac{a}{x+b}+b+c}$, and so $0=a$, which is a contradiction, therefore our assumption that $c\neq -b$ was false. $\endgroup$ – Jonathan Zhou May 4 '18 at 23:45
2
$\begingroup$

I think you need to take into account the fact that the question asks you to prove that $a \neq 0$ before proving that $c=-b$.

I suppose that $x$ is a variable, which will not take the value $x=-b$...

Let's suppose $a=0$.

Then you have $\forall x, x=c$, which is not possible. Hence $a \neq 0$.

Your initial equation becomes, as you said $a(b+c)=(x-c)(x+b)(b+c)$ by multiplying both sides by $\frac{a}{x+b}+c+b$ and then by $x+b$. Both expressions are not null (remember that $x \neq -b$).

Since $a \neq 0$, I have $(b+c)=\frac 1a(x-c)(x+b)(b+c)$ $\forall x$.

Or $(b+c)(\frac 1a(x-c)(x+b)-1)=0$ $\forall x$

This is only possible if $c=-b$.

$\endgroup$
  • $\begingroup$ Why is $x=c$ not possible? Also how did you get from the second last line to the final line? $\endgroup$ – Jonathan Zhou May 4 '18 at 10:24
  • $\begingroup$ @JonathanZhou If $x$ is a variable, it cannot be forced to be $=c$... For the other question, see the edit. $\endgroup$ – Martigan May 4 '18 at 11:11
  • $\begingroup$ Thanks, this solution works nicely too. $\endgroup$ – Jonathan Zhou May 4 '18 at 23:47
1
$\begingroup$

IMO, the claim is wrong. If $a=0$, the equation reduces to $x=c$, unless $c+b=0$. But we know that $c+b=x+b\ne0$.

For example

$$1=\frac0{\dfrac0{1+1}+1+1}+1.$$


As $x+b\ne0$, we can always rewrite

$$x=\frac{a}{\dfrac{a}{x+b}+b+c}+c=\frac{ax+ab}{(b+c)x+b^2+bc+a}+c\\ =\frac{(a+bc+c^2)x+(b^2c+bc^2+ac+ab)}{(b+c)x+b^2+bc+a}.$$

For this to be an identity, the coefficient of $x$ in the denominator must be zero. When this is the case, it simplifies to

$$x=\frac{ax}{a}.$$


In conclusion, the problem statement is at best ambiguous and at worst wrong. It should say "for all $x\ne-b$".

$\endgroup$
  • $\begingroup$ Edited, it did mean for all $x\neq -b$. Thanks. $\endgroup$ – Jonathan Zhou May 4 '18 at 23:36
0
$\begingroup$

Note that from

$$x=\frac{a}{\frac{a}{x+b}+c+b}+c$$

we need

  • $x+b \neq 0 \implies x\neq -b$

and

  • $\frac{a}{x+b}+c+b\neq 0$

then

$$x=\frac{a}{\frac{a}{x+b}+c+b}+c\iff \frac{ax}{x+b}+cx+bx=a+\frac{ac}{x+b}+c^2+bc\\\iff ax+cx^2+bx^2+bcx+b^2x=ax+ab+ac+c^2x+bcx+c^2b+b^2c\\\iff (b+c)x^2+(a+bc+b^2-a-c^2-bc)x-(ab+ac+c^2b+b^2c)=0\\\iff (b+c)x^2+(b^2-c^2)x-(ab+ac+c^2b+b^2c)=0$$

and then for the $x^2$ and $x$ term

  • $b=-c$

and for the constant term

  • $ab+ac+c^2b+b^2c=ab-ab+b^3-b^3=0$
$\endgroup$
  • $\begingroup$ Very good your answers. I have upvoted all users. +1 $\endgroup$ – user401938 May 4 '18 at 9:59
  • $\begingroup$ @Sebastiano Thanks, really appreciate! $\endgroup$ – user May 4 '18 at 10:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.