Question

what is a highest common factor

Answer 1

\[\frac{ 1 }{ 2 }-\frac{ 1 }{ 3 }=\frac{ 1 }{ 5 }\]

Answer 2

of 2,3,5

Answer 3

suppose \[x,y,z\] are positive intergers satisfying the equation
\[\frac{ 1 }{ x }-\frac{ 1 }{ y}=\frac{ 1 }{ z }\] and let \[h\]
be the highest common factorof x,y,z.
Prove that hxyz is a perfect square

Answer 4

HELP

Answer 5

Is "hxyz" just the product of those variables? Sorry, your opening post made me think you were talking about common denominators... a much simpler question, obviously.

Answer 6

yes hxyz is a product

Answer 7

multiplying gives \[yz-xz-xy=0\]

Answer 8

\[h=xyz\]

Answer 9

\[\frac{ 1 }{ x }-\frac{ 1 }{ y }-\frac{ 1 }{ z }=0\]\[\frac{ zy-zx-xy }{ xyz }=0\]

Answer 10

@cwrw238 @dpaInc

Answer 11

I'm doing the same stuff as you on paper... not making too much progress.

you might multiply top and bottom by h to at least get the denominator in the final form of hxyz, but I don't know where that would lead.

Answer 12

I thought about creating some x = hx' type of dummy variables to try to incorporate h as the greatest common factor... didn't know where to take that either, since I just ended up with x', y', and z' added in the mix.

Answer 13

1 is the highest common factor of 2,3,5

Answer 14

you mean every product 4,5,6 : 7,8,6 1 is gdc,i dnt think so

Answer 15

@PaxPolaris @amistre64

Answer 16

\[\frac{xyz}{1}\left(\frac{ 1 }{ x }-\frac{ 1 }{ y}=\frac{ 1 }{ z }\right)\]
\[yz-xz=xy\]
\[z(y-x)=xy\]
\[z=\frac{xy}{y-x}\]
and im assuming x,y,z not equal 0 to begin with since they are all bottom numbers

Answer 17

\[\frac{ 1 }{ x }-\frac{ 1 }{ y}=\frac{ 1 }{ xy/y-x }\]
\[\frac{ 1 }{ x }-\frac{ 1 }{ y}=\frac{y-x }{ xy }\]
\[\frac{ y-x }{ xy }=\frac{y-x }{ xy }\]
pfft, well that might have been the wrong track to take :)

Answer 18

but what does h mean and can we incoperate it in to the equation

Answer 19

gdc(x,y,z)=h

Answer 20

hmm, good point. are we to assume that all these are integer values? are any spose to be positive only? and what is this twisted term "highest common factor"?

Answer 21

the next question goes like prove that \[h(y-z)\].
they are all positive intergers

Answer 22

earlier when you typed h = xyz, was that some definition of h, or something you derived from somewhere?

if h is a greatest common factor, for the case of 1/2 - 1/3 = 1/5, the GCF of those three primes is 1, making h = 1, but 1*2*3*5 = 30, which is not a perfect square. So for that definition of h, this cannot be proved due to this counterexample.

Answer 23

i made a mistake the gdc is not equal to xyz
h is not =xzy