Science Forums 2

[Qfwfq]

Originally Posted by Don Blazys

In post#57, you ended your first paragraph with: "Let's look at the fallacies in your argument".

Well, I already know that there are no fallacies in my argument, so the only possible reply from my point of view was and is: "There are no fallacies in my argument."

Please note that these forums have Rules which include something against this type of reply. The onus of proving your point when challenged is on you and I provided a clear argument against your claim about "reasonable implication" which you had not supported at all.

Originally Posted by Don Blazys

I suppose that I could also have mentioned that "my grasp of fundamental logic" is quite good, and that from my point of view, it's your grasp of fundamental logic that needs to be "straightened out", but I prefer not to indulge in such childish gibberish.

Which childish gibberish?

By definition, an implication is a true assert if there are no cases in which the consequent is false and the implicant true. Without this being the valid requistite, there would be no point in arguing out any implication. It is trivial that it holds if at least one of the following can be said about a given implication:

• there are no cases in which the consequent is false
• there are no cases in which the implicant is true

Now Don, I agree that at if least one of the above holds the implication is useless. Indeed this is the reason I don't see the point of your argument, based on being true. In the second case above, it is impossible to apply modus ponens and superfluous to apply modus tollens; vice versa in the first case. Where am I needing to be straightened out?

Originally Posted by Don Blazys

If the expression:

(1)^(0/0)

was "undefined", as you claim it is, then both (7) and (8) would become:

a^x+b^y=c= "undefined"

and

a^x=b^y=c^2= "undefined",

and that would make no sense whatsoever, because clearly, defined co-prime equations do indeed exist!

The properties of logarithms can not render all co-prime equations "undefined"!

Thus, it must be the case that:

(1)^(0/0)=1,

and

(1)^(0/0) is not "undefined".

Gosh I'm finally beginning to see that your argument is along the lines of:

is an identity (rather than as I say)

therefore, since 1 is defined even when then so must be. This appears to be the main source of disagreement.

Mathematics, as Gauss meant to imply in the sentence you quoted here weeks ago, is all a matter of: "Define, construct and work out the consequences!" Now Don, if you say the above is an identity rather than you imply that the expression defines a function that's equal to 1 and continuous even for the value and therefore, in order to adopt a non-standard definition of the terms function and continuous then you must state exactly what your definition of these terms is. If you can make the whole thing self-consistent it's an Alternative Theory, otherwise it's a Strange Claim and this case includes if you insist on your claims about implications but fail to support them.