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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!
The statement is asking you to prove that raising the product of three real numbers, x, y, and z, to the eighth power will always result in 1.
Proving this statement is important because it demonstrates a fundamental property of real numbers and their operations. It also allows us to understand and apply this property in more complex mathematical problems.
The first step is to rewrite the statement in a more general form, such as $(abc)^n=1$, where n is any integer. This will allow us to use mathematical properties and operations that apply to all real numbers, rather than specific values of x, y, and z.
This proof involves the concepts of exponents, real number operations, and properties of equality. It may also require the use of algebraic manipulation and the properties of the real number system, such as the distributive property and the power of a product rule.
Yes, there is one exception to this statement, which is when at least one of the values of x, y, and z is equal to 0. In this case, the product of the three numbers will also be 0, and raising 0 to any power will result in 0, not 1.