Re: Property of an equation

To Deep: Let me replay some extracts from this thread to try to
isolate just this one issue ...

In a previous thread you gave an example of 2 irrational numbers
which, according to your concept, have "no common factor" (Deep
wrote):

Example: sqrt(3) and 2.sqrt(5) have no factor in common

I replied (quasi wrote):

These conditions need clarification.

... the concept of "no common factor" needs to be specified more clearly.
The example you gave is not clear enough from my perspective.

You then replied (Deep wrote):

Let g/h = m/n (1.c)
If in (1.c) the condition g = m and h = n always hold then g and h
have no factor in common.

Again I objected (quasi wrote):

No, this is still not clear. You're saying that g and h have no common
factor if the equation g/h = m/n implies g=m and h=n. That doesn't
make sense to me. For example, certainly g/h = (2g)/(2h).

Before you can talk meaningfully about common factors, you need to
specify the set of all possible factors. If g,h were restricted to a
given ring, then perhaps the concept could be made precise. Even then,
you would have to worry about whether factorization in your ring is
always finite and if it is, whether you have unique factorization.

In your reply, perhaps you thought that you were clarifying, but
instead you reiterated the same flawed concept (Deep wrote):

Let me again clarify what is a "common factor". Let g ,h, r, m,. n be
real numbers. No other restrictios are imposed.
Let g = rm and h = rn. If g/h = m/n then r is the common factor.
Otherwise, g and h have no factor in common. The case r= 1 need not be
considered.

I replied (quasi wrote):

No, you can't get away with the above concept of common factor, so
slow down and let's try to resolve the issue. You can't just brush it
aside.

Your "definition" of common factor for real numbers is seriously
flawed. Let g and h be any two reals numbers. Then, by your
specification, _every_ nonzero real number r would be a common factor
of g and h. Simply let m=g/r and n=h/r and you get g=rm and h=rn.
Thus, unless you revise the concept of what you mean by common factor,
you can't produce a pair of real numbers which actually have a
greatest common factor.

You never responded to substance of the above objection. Instead, in
your next reply, you make a still more general claim, based on the
same flawed concept of common factor (Deep wrote):

Now based on my previous explanations consider (1.2) and
(2.2) as below.

X = (g^2)(G^2) (1.2); Y = (h^2)(H^2)
(2.2)

X, Y are nonsquare integers each > 1 and none is a prime. There is no
common factor between g^2 and G^2, No common factor between h^2 and
H^2; (X, Y) = 1.

Under these circumstances I conclude, g^2 is an integer and G^2 is
also an integer.

If you disagree with my conclusion kindly tell me your reason of
disagreement so that I can adequately address your issue.
Otherwise, I would believe that you agree with my conclusion.

I then replied (quasi wrote):

I don't agree with it. You first have to resolve the meaning of common
factor for pairs of real numbers. You show me two real numbers which
you think have no common factor and I'll show you that you're wrong
(unless you either revise your set of allowable real numbers, or
revise your definition of common factor, or both).

In your latest reply, you continue to ignore the issue of common
factors, asking instead only whether or not I agree with your claim
about g^2 and h^2 (Deep wrote):

My assertion: both g^2 and h^2 are positive real integers.
...
If you donot agree with the assertion, kindly let mev know why?. I wil
then respond to you
in more meaningful way.

There's no point in discussing your claim about g^2 and h^2. You need
to put aside the question of whether or not your equations imply some
conclusion or other, and first deal with the issue of what you mean by
factors of real numbers. It's not just a minor detail -- it's a
fundamental objection which blocks all further analysis until the
issue is resolved.

quasi
.

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