~True or False??

[bomba923]

See the attached image for the question :)

[Hustler]

See the attached image for the question :)

True. Isn't this trivial?

[Dr. Haight]

I'll say False, just because voting the opposite of Hustler will, in most cases, end in my advantage.

[bomba923]

True. Isn't this trivial?

It's trivial for q=1. But what if q > 1 ? ;)
(Is the statement still true?)

[Hustler]

It's trivial for q=1. But what if q > 1 ? ;)
(Is the statement still true?)

I'm going to say yes, it's still true. It's also trivial in the case that r <= p.

[PiccoloNamek]

Wow, I think my brain just exploded.

[indie]

t's trivial for q=1. But what if q > 1 ?
(Is the statement still true?)

I'm always annoyed by the "p > q" assumption for set equations . . . p comes before q in the alphabet, and therefore has a lower ordinal value. I tend to think ordinal before cardinal usually, intuitively.

However, I can see how it might be relevent for something like a code for "p" = problem and "q" = question, where the problem is larger than the question, which is clearly larger than zero, and in the "set" which is either inclusive or exclusive of a letter paired with a subscript of {0,1 . . . ,n} as a summation of the letters' relationship to cardinal numbers. Likewise , "p" could = prime and "q" = quotient of some strange relationship that is not entirely trivial. Or any one of other various enigmas.

But anyway. If q > 1, I would guess that the answer would be False

If the assumption is that p > q > 0
and it's stated that q > 1,
p cannot be greater than itself subract one when it's a given that q > 0

But I just like to formulate strange answers to intriguing questions sometimes; usually I don't know what I'm talking about. ;)

[bomba923]

I'm going to say yes, it's still true. It's also trivial in the case that r <= p.

How do you know it is true? ALL you've shown are just two trivial cases //

Right, it is trivially true in the specific case where q=1 or r <= p :rolleyes2
(But how do you know it is true if q > 1 and r > p ? )

In other words, what about the statement IN GENERAL? ;)
For all possible cases (trivial and non-trivial),

Is the statement true or false?

[Ferrus]

Now I remember why I gave up Maths.

[Hustler]

Right, it is trivially true in the specific case where q=1 or r <= p :rolleyes2
(But how do you know it is true if q > 1 and r > p ? )

In other words, what about the statement IN GENERAL? ;)
For all possible cases (trivial and non-trivial)

Yeah, I know. But that's the hard part. I'm just guessing it's true, but that's only because I can't come up with a counterexample.

[euterpenc]

Clearly math is inadequate as that amounts to little more than a bunch of symbols, with no real meaning.

[KuJo]

Clearly math is inadequate as that amounts to little more than a bunch of symbols, with no real meaning.

yes, well it has more meaning than you. and by your definition you shouldnt be alive.