March 5, 2011 10:48 AM Subscribe

I'm trying to express my love for a certain dessert in a math/logic formula.
I have this: **π > ∼π **, which I take to mean "Pi is greater than not Pi", and this: **π > ∞-π**, which I take to mean "Pi is greater than Everything but Pi". Do these make any sense or hold up in any legit way?

posted by TheCoug to Food & Drink (15 answers total) 5 users marked this as a favorite

posted by TheCoug to Food & Drink (15 answers total) 5 users marked this as a favorite

Pretty much not. The “not” symbol is for Boolean logic, which doesn’t have a concept of “greater than”. And pi is most certainly not greater than an infinite quantity minus pi, which is still infinity.

If you wanted “pi is greater than not pi”, you could say π > 0π, since 0 multiplied by anything is 0, and π is most certainly greater than 0. Here 0π is “the absence of pi”.

For “pi is greater than everything but pi”, I don’t think there really is a concept that fits. “Everything but pi” is conceptually closer to a set than a number, and sets themselves aren’t automatically comparable in terms of “greater than” and “less than”. (The relevant concept here is the existence (or lack thereof) of a well-ordering for a set of sets of numbers.

On preview: if you just wanted things that use the correct symbols but weren’t actually true, go with what grouse said.

posted by spitefulcrow at 11:08 AM on March 5, 2011

If you wanted “pi is greater than not pi”, you could say π > 0π, since 0 multiplied by anything is 0, and π is most certainly greater than 0. Here 0π is “the absence of pi”.

For “pi is greater than everything but pi”, I don’t think there really is a concept that fits. “Everything but pi” is conceptually closer to a set than a number, and sets themselves aren’t automatically comparable in terms of “greater than” and “less than”. (The relevant concept here is the existence (or lack thereof) of a well-ordering for a set of sets of numbers.

On preview: if you just wanted things that use the correct symbols but weren’t actually true, go with what grouse said.

posted by spitefulcrow at 11:08 AM on March 5, 2011

On what should have been preview: “an infinite quantity minus pi” is still “an infinite quantity”, not “infinity”.

posted by spitefulcrow at 11:09 AM on March 5, 2011

posted by spitefulcrow at 11:09 AM on March 5, 2011

∀ x, x + π > x + 0π

for all x, x plus pi > x plus no pi

rough translation: everything is improved by pie

posted by robtoo at 11:26 AM on March 5, 2011 [6 favorites]

for all x, x plus pi > x plus no pi

rough translation: everything is improved by pie

posted by robtoo at 11:26 AM on March 5, 2011 [6 favorites]

My answers obviously aren't true for the real numbers. They presupposes a partially ordered set where π is the greatest element. Including an element that is the set of all of the other elements, as is done in my second equation, seems like a contradiction, though.

posted by grouse at 11:34 AM on March 5, 2011

posted by grouse at 11:34 AM on March 5, 2011

Other versions:

∀ x. π > x ("pie is greater than everything" - suffers from the awkward "is pie greater than itself?")

∀ x. x ≠ π → π > x (improved version of grouse's first formula above, makes explicit we mean it for eveythring that is not pie - in absence of quantifiers, one can't really claim so in a technical sense, though it IS standard mathematical practice to assume variables are universally quantified so this is mostly nitpicking)

∀ x,y. Pie(x) ∧ ¬Pie(y) → x>y (for any x that is pie and any y that is not, x is greater than y)

posted by Iosephus at 11:42 AM on March 5, 2011

∀ x. π > x ("pie is greater than everything" - suffers from the awkward "is pie greater than itself?")

∀ x. x ≠ π → π > x (improved version of grouse's first formula above, makes explicit we mean it for eveythring that is not pie - in absence of quantifiers, one can't really claim so in a technical sense, though it IS standard mathematical practice to assume variables are universally quantified so this is mostly nitpicking)

∀ x,y. Pie(x) ∧ ¬Pie(y) → x>y (for any x that is pie and any y that is not, x is greater than y)

posted by Iosephus at 11:42 AM on March 5, 2011

My stab at "pi is greater than everything that is not pi": ∀ x ∈ {x | x ≠ π}, π > x

Maybe a little plodding and redundant, though.

posted by en forme de poire at 11:44 AM on March 5, 2011

Maybe a little plodding and redundant, though.

posted by en forme de poire at 11:44 AM on March 5, 2011

Another option: ∀x ∈ {x : x ≠ π}, x < π.

This is equivalent to the first expression grouse supplied. Of course, without an explicit link between π and pie, this is nonsense. Might want to replace π with the image of a pie, you know, to make sure you're talking about some set of things that includes pie, rather than just the set of real numbers.

posted by Nomyte at 11:44 AM on March 5, 2011

This is equivalent to the first expression grouse supplied. Of course, without an explicit link between π and pie, this is nonsense. Might want to replace π with the image of a pie, you know, to make sure you're talking about some set of things that includes pie, rather than just the set of real numbers.

posted by Nomyte at 11:44 AM on March 5, 2011

Thanks guys, I think that gives me enough to go on, but feel free to keep 'em coming!

posted by TheCoug at 11:46 AM on March 5, 2011

posted by TheCoug at 11:46 AM on March 5, 2011

Risking being annoying by now (professional deformation, heh), yes there is a risk that the logic one works in falls into paradoxes when everything works at the same level (here, elements and sets of elements). Solving this typically involves some kind of type theory so the levels get separated.

And yes, while ditching the pi symbol would diminish the joke, more logic representations of your notions can be found in the style of my last formula. In fact, this is the most straightforward way one goes around doing KR ("knowledge representation") in computer science. In such work, one even gets insufferable and does things like:

∀ x,y. Pie(x) ∧ Food(y) ∧ ¬Pie(y) → x>y

if one wants to say that pie is the greatest food of all, but not that pie is, say, greater than kittens, movies, or Mt Vesuvius.

posted by Iosephus at 11:55 AM on March 5, 2011

And yes, while ditching the pi symbol would diminish the joke, more logic representations of your notions can be found in the style of my last formula. In fact, this is the most straightforward way one goes around doing KR ("knowledge representation") in computer science. In such work, one even gets insufferable and does things like:

∀ x,y. Pie(x) ∧ Food(y) ∧ ¬Pie(y) → x>y

if one wants to say that pie is the greatest food of all, but not that pie is, say, greater than kittens, movies, or Mt Vesuvius.

posted by Iosephus at 11:55 AM on March 5, 2011

Iosephus:

I think grouse said it best when correctly presupposed that I was living in an unordered set where Pie is the greatest element. I mean, it's just the greatest!

posted by TheCoug at 12:01 PM on March 5, 2011

I think grouse said it best when correctly presupposed that I was living in an unordered set where Pie is the greatest element. I mean, it's just the greatest!

posted by TheCoug at 12:01 PM on March 5, 2011

Partially ordered set. Unordered means there is no greatest element.

robtoo's formulation has the advantage that it actually makes sense with the standard conception of π and real numbers, and requires less advanced math.

posted by grouse at 12:12 PM on March 5, 2011

robtoo's formulation has the advantage that it actually makes sense with the standard conception of π and real numbers, and requires less advanced math.

posted by grouse at 12:12 PM on March 5, 2011

There's a complication... if (F, ≤) is the set of foods (as represented by real numbers) partially-ordered by greatness, and π = max(F), then we have the problem that π < sup(F), where sup is the supremum (least upper bound). This could be interpreted to mean that pie is not as good as a soup made out of all possible foods. While we can't actually check this, it doesn't seem like a very desirable property of our theory. There's also the problem that sup(F) is not in F - i.e. said soup is not a food. In culinary terms this may be reasonable, but mathematically it is counter-intuitive at best.

To fix this, I suggest that we allow combinations of foods to be considered separately - so that π < 2π < 3π < ... as we'd expect. Two pies are clearly better than one so this may in fact be a better model. The downside is that soup no longer exists, and a single pie is not the greatest any more. But thanks to Archimedes, we at least have that for any collection of food you care to take, you can do better simply by using a large enough number of pies.

Hope that helps.

posted by d11 at 12:18 PM on March 5, 2011 [3 favorites]

To fix this, I suggest that we allow combinations of foods to be considered separately - so that π < 2π < 3π < ... as we'd expect. Two pies are clearly better than one so this may in fact be a better model. The downside is that soup no longer exists, and a single pie is not the greatest any more. But thanks to Archimedes, we at least have that for any collection of food you care to take, you can do better simply by using a large enough number of pies.

Hope that helps.

posted by d11 at 12:18 PM on March 5, 2011 [3 favorites]

d11, I think that is not correct. If my memory serves, any subset of the reals that has a greatest element has that same element as the supremum of the subset.

posted by Iosephus at 12:51 PM on March 5, 2011 [1 favorite]

posted by Iosephus at 12:51 PM on March 5, 2011 [1 favorite]

(∀ q : (q ∉ J{ π })) ∧ (∀ p : (p ∈ J{ π })) • p > q

Translation:

for all q such that q is not a member of the set of all sets {pi}

and

for all p such that p is a member of the set of all sets {pi}

it holds that p is greater than q

or to put it another way

for all things q that are not in the set of things that are pie,

and

for all things p that are in the set of things that are pie,

p is greater than q

posted by tel3path at 1:45 PM on March 5, 2011

Translation:

for all q such that q is not a member of the set of all sets {pi}

and

for all p such that p is a member of the set of all sets {pi}

it holds that p is greater than q

or to put it another way

for all things q that are not in the set of things that are pie,

and

for all things p that are in the set of things that are pie,

p is greater than q

posted by tel3path at 1:45 PM on March 5, 2011

This thread is closed to new comments.

This means that π is greater than anything that is not π:

x \neq \pi \Rightarrow \pi > x

This means that π is greater than all the things that are not π combined:

\pi > \bigcup_{x \neq \pi} x

posted by grouse at 11:05 AM on March 5, 2011