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#### Albert

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- Jan 25, 2013

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$x,y,z \,\, \in R$

$x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}$

$prove :x^8y^8z^8=1$

$x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}$

$prove :x^8y^8z^8=1$

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- Jan 25, 2013

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$x,y,z \,\, \in R$

$x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}$

$prove :x^8y^8z^8=1$

$x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}$

$prove :x^8y^8z^8=1$

- Jan 26, 2012

- 183

$x + \dfrac{1}{y} = y + \dfrac{1}{z} = z + \dfrac{1}{z} = 2 \dfrac{1}{2}$

yet $x^8y^8z^8 \ne 1$.

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- Jan 25, 2013

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if x=y=z then this problem doesn't make any sense.

$x + \dfrac{1}{y} = y + \dfrac{1}{z} = z + \dfrac{1}{z} = 2 \dfrac{1}{2}$

yet $x^8y^8z^8 \ne 1$.

of course here x=y=z will be excluded

- Jan 26, 2012

- 183

And why does $x=y=z$ not make sense?

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- Jan 25, 2013

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$x-y=\dfrac {1}{z}-\dfrac {1}{y}=\dfrac {y-z}{yz}$$x,y,z \,\, \in R$

$x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}$

$prove :x^8y^8z^8=1$

$\therefore yz=\dfrac {y-z}{x-y} ,( here \,\, x\neq y)------(1)$

$\text {likewise :}$

$ xy=\dfrac {x-y}{z-x} ,( here \,\, z\neq x)------(2)$

$ zx=\dfrac {z-x}{y-z} ,( here \,\, y\neq z)------(3)$

$(1)\times (2)\times (3) :x^2y^2z^2=1$

$\therefore x^8y^8z^8=1$

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- Jan 26, 2012

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$x^{8}y^{8}z^{8}=1$ seems a very odd conclusion for such a theorem. Normally, a conclusion like that would be for the purpose of illustrating why the equality$x-y=\dfrac {1}{z}-\dfrac {1}{y}=\dfrac {y-z}{yz}$

$\therefore yz=\dfrac {y-z}{x-y} ,( here \,\, x\neq y)------(1)$

$\text {likewise :}$

$ xy=\dfrac {x-y}{z-x} ,( here \,\, z\neq x)------(2)$

$ zx=\dfrac {z-x}{y-z} ,( here \,\, y\neq z)------(3)$

$(1)\times (2)\times (3) :x^2y^2z^2=1$

$\therefore x^8y^8z^8=1$

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- Jan 25, 2013

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$ \text{since} \,\, x,y,z\in R$$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 equalitydoesn'twork for a lower power, rather like why $x^{2}+y^{2}$ does not factor over the reals, but $x^{4}+y^{4}$ does.

$x^2y^2z^2=1$

$\text{implies}\,\,\, xyz= \pm 1$

$\therefore x^8y^8z^8=1$

- Jan 26, 2012

- 183

leads to$x-y=\dfrac {1}{z}-\dfrac {1}{y}=\dfrac {y-z}{yz}$

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.$\therefore yz=\dfrac {y-z}{x-y} $

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- Jan 25, 2013

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yes the original statement is true (like 2=2=2 ,or 3=3=3)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.

but the statement is otiose,and as you said will not give to my result

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- Jan 26, 2012

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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 $(xyz)^{8}=1$.$ \text{since} \,\, x,y,z\in R$

$x^2y^2z^2=1$

$\text{implies}\,\,\, xyz= \pm 1$

$\therefore x^8y^8z^8=1$

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- #11

- Jan 25, 2013

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$ginen \,\, x,y,z \in R$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 $(xyz)^{8}=1$.

statement (1)

$if \,\, (xyz)^{2}=1 \,\, then \,\, (xyz)^{2k}=1( k\in N)$

statement (2)

$if \,\, (xyz)^{2k}=1 \,\, then \,\, (xyz)^{2}=1( k\in N)$

these two statements which one is true ?

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- #12

- Jan 26, 2012

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In the real number system, they would be exactly equivalent. However, if you allow complex numbers into the picture, the first is true and the second is not. Counterexample: $(xyz)^{4}=1$ but $(xyz)^{2}=-1$.$ginen \,\, x,y,z \in R$

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 ?

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- #13

- Jan 25, 2013

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so as you said In the real number system, they would be exactly equivalent (statementIn the real number system, they would be exactly equivalent.

if you allow complex numbers into the picture, the first is true

(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 ?

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- #14

- Jan 26, 2012

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I don't think I'm communicating very well here. It's not that I object to theso 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 ?

I object to the

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

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- #15

- Jan 25, 2013

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Ok! taking aesthetics into consideration I agree with you the following work is more beautifulTherefore, 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'mnotsaying that either one of themfails to bea work of art!

it may sell better price in an auction market (besides it is easier for students at first glance)

Given that $x+1/y=y+1/z=z+1/x$, and $x\not=y, y\not=z, z\not=x$, prove that $(xyz)^{2}=1$.