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Prove (xyz)^8=1

Albert

Well-known member
Jan 25, 2013
1,225
$x,y,z \,\, \in R$

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

$prove :x^8y^8z^8=1$
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
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

Well-known member
Jan 25, 2013
1,225
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 $x^8y^8z^8 \ne 1$.
if x=y=z then this problem doesn't make any sense.

of course here x=y=z will be excluded
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
And why does $x=y=z$ not make sense?
 

Albert

Well-known member
Jan 25, 2013
1,225
$x,y,z \,\, \in R$

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

$prove :x^8y^8z^8=1$
$x-y=\dfrac {1}{z}-\dfrac {1}{y}=\dfrac {y-z}{yz}$

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

$\text {likewise :}$

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

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

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

$\therefore x^8y^8z^8=1$
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
$x-y=\dfrac {1}{z}-\dfrac {1}{y}=\dfrac {y-z}{yz}$

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

$\text {likewise :}$

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

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

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

$\therefore x^8y^8z^8=1$
$x^{8}y^{8}z^{8}=1$ 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.
 

Albert

Well-known member
Jan 25, 2013
1,225
$x^{8}y^{8}z^{8}=1$ 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.
$ \text{since} \,\, x,y,z\in R$

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


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

$\therefore x^8y^8z^8=1$
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
I understand that
$x-y=\dfrac {1}{z}-\dfrac {1}{y}=\dfrac {y-z}{yz}$
leads to

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

Well-known member
Jan 25, 2013
1,225
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.
yes the original statement is true (like 2=2=2 ,or 3=3=3)

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

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
$ \text{since} \,\, x,y,z\in R$

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


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

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

Well-known member
Jan 25, 2013
1,225
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 $(xyz)^{8}=1$.
$ginen \,\, x,y,z \in R$

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

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

these two statements which one is true ?
 
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Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
$ginen \,\, x,y,z \in R$

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 ?
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 $(xyz)^{2}=-1$.
 

Albert

Well-known member
Jan 25, 2013
1,225
In the real number system, they would be exactly equivalent.
if you allow complex numbers into the picture, the first is true
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 ?
 
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Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
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 ?
I don't think I'm communicating very well here. It's not that I object to the truthfulness of this challenge problem. I would agree that if $x+1/y=y+1/z=z+1/x$, and if $x\not=y$ and $y\not=z$ and $z\not=x$, then $(xyz)^{8}=1$. There's no need even to insist on $x,y,z\in \mathbb{R}$ - they could be complex. You proved this in post 5. So I'm not objecting to whether your challenge problem is correct or not.

I object to the aesthetics of this challenge problem. In post 5, you proved something stronger than $(xyz)^{8}=1$, namely, $(xyz)^{2}=1$. 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!
 
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Albert

Well-known member
Jan 25, 2013
1,225
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!
Ok! taking aesthetics into consideration I agree with you the following work is more beautiful
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 $(xyz)^{2}=1$.