Albert1
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$x,y,z \,\, \in R$
$x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}$
$prove :x^8y^8z^8=1$
$x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}$
$prove :x^8y^8z^8=1$
Jester said:Something doesn't look right. Consider $x = y= z = 2$. Then clearly
$x + \dfrac{1}{y} = y + \dfrac{1}{z} = z + \dfrac{1}{z} = 2 \dfrac{1}{2}$
yet $x^8y^8z^8 \ne 1$.
Albert said:$x,y,z \,\, \in R$
$x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}$
$prove :x^8y^8z^8=1$
Albert said:$x-y=\dfrac {1}{z}-\dfrac {1}{y}=\dfrac {y-z}{yz}$
$\therefore yz=\dfrac {y-z}{x-y} ,( here \,\, x\neq y)------(1)$
$\text {likewise :}$
$ xy=\dfrac {x-y}{z-x} ,( here \,\, z\neq x)------(2)$
$ zx=\dfrac {z-x}{y-z} ,( here \,\, y\neq z)------(3)$
$(1)\times (2)\times (3) :x^2y^2z^2=1$
$\therefore x^8y^8z^8=1$
$ \text{since} \,\, x,y,z\in R$Ackbach said:$x^{8}y^{8}z^{8}=1$ seems a very odd conclusion for such a theorem. Normally, a conclusion like that would be for the purpose of illustrating why the equality doesn't work for a lower power, rather like why $x^{2}+y^{2}$ does not factor over the reals, but $x^{4}+y^{4}$ does.
leads toAlbert said:$x-y=\dfrac {1}{z}-\dfrac {1}{y}=\dfrac {y-z}{yz}$
only if $x \ne y$ but the first statement is still true if $x=y=z$ so $x=y=z$ makes sense, just not to give your result.Albert said:$\therefore yz=\dfrac {y-z}{x-y} $
Jester said:I understand that leads to only if $x \ne y$ but the first statement is still true if $x=y=z$ so $x=y=z$ makes sense, just not to give your result.
Albert said:$ \text{since} \,\, x,y,z\in R$
$x^2y^2z^2=1$$\text{implies}\,\,\, xyz= \pm 1$
$\therefore x^8y^8z^8=1$
$ginen \,\, x,y,z \in R$Ackbach said:I understand that. My comment was intended to say that, as mathematical aesthetics go, since $x^{2}y^{2}z^{2}=1$ is a stronger conclusion than $x^{8}y^{8}z^{8}=1$, you would generally want to stop at the smaller power. What I'm saying is that I think the problem should be to prove $(xyz)^{2}=1$, not $(xyz)^{8}=1$.
Albert said:$ginen \,\, x,y,z \in R$
statement (1)
$if \,\, (xyz)^{2}=1 \,\, then \,\, (xyz)^{2k}=1( k\in N)$
statement (2)
$if \,\, (xyz)^{2k}=1 \,\, then \,\, (xyz)^{2}=1( k\in N)$
thes two statements which one is true ?
so as you said In the real number system, they would be exactly equivalent (statementAckbach said:In the real number system, they would be exactly equivalent.
if you allow complex numbers into the picture, the first is true
Albert said:so as you said In the real number system, they would be exactly equivalent (statement
(1) and statement(2) )
$x,y,z \in R \,\, and \,\, k\in N$
$ if \,\, (xyz)^2=1 \,\, then \,\, (xyz)^{2k}=1$ ,---- statement(1)
$\therefore if \,\, (xyz)^2=1 \,\, then \,\, (xyz)^8=1$
(even we allow complex numbers into the picture, the first is true,but here we only take real
numbers into consideration)
now any objection ?
Ackbach said:Therefore, I would argue that posing the following challenge problem is more aesthetically pleasing:
Given that $x+1/y=y+1/z=z+1/x$, and $x\not=y, y\not=z, z\not=x$, prove that $(xyz)^{2}=1$.
So, to reiterate, what I'm saying here is like saying that one work of art is more beautiful than another. I'm not saying that either one of them fails to be a work of art!