## [2014-09-27] Math Party

Blame Quintushalls for this.

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Peon
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### [2014-09-27] Math Party

http://www.smbc-comics.com/?id=3495#comic

I'm not a big fan of this one because the terms used in the proof are a bit handwavy and i've never seen a proof by induction count down to one like that. The whole idea of inductive proofs is that you show it's true for an infinite list of things, but clearly a group of people in a house is finite. Am i even sadder than the man in the comic because i see the proof is flawed? Alternatively do any of you know of a proof by induction that counts down like that? I feel like zach was just being sloppy.
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Zweisteine

### Re: [2014-09-27] Math Party

It's the Sorites Paradox.

Typical version:
You have a heap of sand. If you remove a single grain, it is still a heap. ...... Therefore, a single grain of sand must be a heap.

I don't want to type a long rant about it, so read the wikipedia page (linked above) if you want more.

Guest

### Re: [2014-09-27] Math Party

This does not follow a legitimate proof by induction. In fact, it's practically a proof by assertion, because the number it starts with is arbitrary and unstated. It's begging the question rather than inductive proof.

Gangler
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### Re: [2014-09-27] Math Party

Mostly it just seems to point out that language sometimes conveys concepts that intuitively make sense but don't really hold up to precise measurements. Which I imagine is why more precise endeavors like cooking tend to avoid terminology like "A heap of Nackle" or "A party of shrimp".

But for less precise endeavors you don't really need to know the exact number at which a party ceases to be a party, so long as everyone can agree on approximately how many people is appropriate for the event in question. It can be contextually sensitive too, allowing a small gathering of friends in the basement to co-exist with a communal bash in the streets under the same label.
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Dinosaur

### Re: [2014-09-27] Math Party

For me, there is a glaring reason why this fails to be mathematical induction: it postulates no base case. So yeah, you get absurd things like a 'party' containing one, zero, or even a negative number of attendees.

Zweisteine

### Re: [2014-09-27] Math Party

Yes, but as stated in the Wikipedia article, you can supply numbers, but it is difficult to determine at what point the pile is no longer a heap.

If a heap is defined as 10,000 or more grains of sand, it makes little sense (by observation and logic, not math) to say 9,999 grains of sand is not a heap. This causes issues that the wikipedia article probably talks about more (not much more than I do).

One solution from the article is to define a heap as any number of grains of sand in which some grains are supported entirely by other grains. This leads to some very small heaps, but it works. Applying this to this comic, you simply have to have a better definition of party than "if you have a group of people at your house." The man's mistake was in the first panel of the comic.

The issue is that this "paradox" attempts to apply logical ideas to illogical concepts. A heap or a party is not an item defined by math, it is a concept based on what you observe. Mixing this with math-y proofs doesn't work.

I do agree that this does't really hold up mathematically, but it's an interesting idea to think about for a few minutes (but not longer).

Peon
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### Re: [2014-09-27] Math Party

To be clear, the heap/nonheap or party/nonparty thing is interesting, but it is specifically the reference to induction that I take issue with, and i'm pretty certain it's incorrect. So the strip is bad
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Malky

### Re: [2014-09-27] Math Party

induction only works if:

1) s(k) + K = s(k+1) is true and

2) s(1) is true.

Kaharz
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### Re: [2014-09-27] Math Party

I too don't think it is a proof by induction, because it is a joke.
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Peon
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### Re: [2014-09-27] Math Party

Shit joke.
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DonRetrasado
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### Re: [2014-09-27] Math Party

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AmagicalFishy

### Re: [2014-09-27] Math Party

Let n be a positive integer. Let Q(n) be the statement: A group of 51 - n people is a party.

The Base Case
Q(1) surely holds true. A group of 50 people is a party.

The Induction Step
Assume Q(m) is true for all m. That is, a group of m people is a party. Surely, removing a single person from a party will not alter it's party status. Thus, if Q(m) is true, then Q(m + 1) is true.

Since the base case holds true for n = 1 and Q(m) implies Q(m+1), the statement Q(n) is true for all n > 1. Therefore, one person is a party where n = 50.

QED.

The problem isn't the induction, nor is it needing to have a base-case begin at 1 (which isn't a necessity, I just did it because your mother wears combat boots). The problem is trying to apply rigorous mathematical methods to vague and totally arbitrary concepts.

Peon
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### Re: [2014-09-27] Math Party

You seem to be conveniently ignoring cases where n is 51 or greater, despite claiming in your "proof" that it works for all n > 1.
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Lethal Interjection
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### Re: [2014-09-27] Math Party

Let's let bygones be "Bye!" Gone.

AmagicalFishy

### Re: [2014-09-27] Math Party

Peon wrote:You seem to be conveniently ignoring cases where n is 51 or greater, despite claiming in your "proof" that it works for all n > 1.
AmagicalFishy wrote:The problem isn't the induction, nor is it needing to have a base-case begin at 1 (which isn't a necessity, I just did it because your mother wears combat boots). The problem is trying to apply rigorous mathematical methods to vague and totally arbitrary concepts.
that's your mother, specifically

When n = 51, it is implied that -1 people is a party.

We know it's true for 1. We also know that if it's true for n, it's true for n+1. Thus, let n = 1. This means we know it's true for 2. Now we also know it's true for 3—and 4, and 5, and 6, and 7, and 8, and 9, and 10, and 11, and 12, and 13, and 14, and 15, and 16, and 17, and 18, and 19, and 20, and 21, and 22, and 23, and 24 and ... and 51. This is the power of induction. If you successfully prove something that is well defined by induction, you've successfully proven it. There are no caveats or anything—it's a solid, water-tight proof.

So, I'm not ignoring the case (and am thus doing nothing convenient!!)—it is implied by n = 1 being true. My proof is solid (except for the fact that it's applied to vague and totally arbitrary concepts). Vague and totally arbitrary concepts ≠ well-defined. The induction method isn't the problem, it's what it's being applied to.

(Don't get me wrong: This doesn't mean we can't apply induction to non-math things. They just have to be well defined. For example, say I have a pitri dish with a single bacterium which, over time, multiplied into a whole colony. This bacteriam had Gene A. Let's index all bacteria in my pitri dish with the counting numbers (1 to ) in the order of their spawning; if n spawned at the same time, choose the n next numbers arbitrarily for each—but the first bacterium is always 1, and all other numbers came from that bacterium.

Now, say we have some biological law that if a bacterium has Gene A, then all bacteria which spawn directly from it have Gene A. We can use induction to show that all bacteria in my pitri dish must have Gene A (even though they all didn't spawn directly from the first one). I'm not going to write out the proof, but I imagine the idea seems pretty intuitive. Of course all bacteria have Gene A.

But that's all induction is—it's that same deductive (lol) logic, but rigorously math.)