Math Is Fun Forum

It doesn't take much time to post 1-3 word replies to threads which add nothing to the discussion, such as "cool", "ok", "lol that's funny", and so on.

Why does post count matter so much to you people?

22/7 is Pi Approximation Day! Funny that 3.14 is an approximation as well, yet it still is considered the "official" pi day...

The answer is "Why not?"

I miss the dinosaurs

Bigger is better, enlarge the banner please

Yes, it can be considered an effect of the completeness property of the reals: every nonempty set of real numbers with an upper bound has a supremum in R. The set S I used is bounded above since every element is less than or equal to 1 (don't think I am implying that 1 must be in the set here, although I did prove it, this is just the definition of an upper bound: for any s in S, s ≤ 1, the ≤ is in the definition. In general the upper bound is not necessarily an element of the set), and in fact the supremum is 1. A property of the supremum is that for any r > 0, there is an s[sub]r[/sub] ∈ S such that sup S - r < s[sub]r[/sub]. Since sup S is 1 in our case, we have that 1 - r < s[sub]r[/sub] ≤ 1. This means s[sub]r[/sub] ∈ B(1, r) = (1 - r, 1 + r). In case you need this defined, an accumulation point (also called a limit point) of a set A is a point x such that every open ball B(x, r) contains a point in A distinct from x for any radius r. We see from the argument several sentences ago that 1 is by definition an accumulation point of S. There is another theorem of point set topology that states a closed set contains all of its accumulation points. This is why it follows that 1 ∈ S. If you need any of the theorems I am referencing explicitly stated and proven, just ask.

I can see why someone might argue that 0.9 + 0.09 + 0.009 + ... only approaches 1. But with basic point set topology, we can easily prove that it actually is 1.

Theorem: Let S = {0.9, 0.99, 0.999, ...}. Then 1 ∈ S.

Proof: First, note that S is closed, since its complement R\S = (-∞, 0.9) U (0.99, 0.999) U ... is a union of open sets, and thus open. Also, 1 is an accumulation point (limit point) of S, since any open ball B(1, r) contains some point in S that is not equal to 1 (I have not seen anyone dispute that 0.9 + 0.09 + 0.009 + ... approaches 1, and this is basically what this sentence says). But S is closed, and thus contains all of its accumulation points. Then 1 ∈ S.

It should be clear that 0.999... is the element of S that is equal to 1.

One time when I was a young boy my cat went insane and jumped on my head, tearing up my scalp for a good while. I still love cats though . Cats are so nice to have hang out with you when they just sit on your lap or bed and relax.

Then what is 0.999...? How about 0.333...? Is pi not a number? Define a number in such a way that 0.999... is not a number but all our notions of what a number is remain intact.

In my proof it is clear that the n appearing in 1/10[sup]n[/sup] and r[sub]n[/sub] is a natural number. I'm sure you know that ∞ is not a natural number, much less a reach number. Then there is no 1/10[sup]∞[/sup] or r[sub]∞[/sub]. n just increases indefinitely, is it so wrong to count 1, 2, 3, 4... forever? You'll always be saying a natural number.

I can rewrite my proof to give another infinite decimal representation for a real number, a representation you wouldn't argue against. This proof is equally as valid as the original; it merely gives another decimal representation for a real number, the "standard" decimal representation.

Theorem. Assume x ≥ 0. Then for every integer n ≥ 1 there is a finite decimal r[sub]n[/sub] = a[sub]0[/sub].a[sub]1[/sub]a[sub]2[/sub]...a[sub]n[/sub] such that r[sub]n[/sub] ≤ x < r[sub]n[/sub] + 1/10[sup]n[/sup].

Proof. Let S be the set of all nonnegative integers less than x. Then S is nonempty, since 0 ∈ S, and S is bounded above by x. Then by the Completeness Axiom of the real numbers, S has a supremum, which we denote a[sub]0[/sub] = sup S. a[sub]0[/sub] ∈ S, so that a[sub]0[/sub] is a nonnegative integer. a[sub]0[/sub] is then the greatest integer in x, [x]. Clearly, a[sub]0[/sub] ≤ x < a[sub]0[/sub] + 1. Now let a[sub]1[/sub] = [10x - 10a[sub]0[/sub]], the greatest integer in 10x - 10a[sub]0[/sub]. Since 0 ≤ 10x - 10a[sub]0[/sub] = 10(x - a[sub]0[/sub]) < 10, we have that 0 ≤ a[sub]1[/sub] ≤ 9 and a[sub]1[/sub] ≤ 10x - 10a[sub]0[/sub] < a[sub]1[/sub] + 1. Then a[sub]1[/sub] is the largest integer satisfying a[sub]0[/sub] + a[sub]1[/sub]/10 ≤ x < a[sub]0[/sub] + (a[sub]1[/sub] + 1)/10. More generally, having chosen a[sub]1[/sub],..., a[sub]n-1[/sub] with 0 ≤ a[sub]i[/sub] ≤ 9, let a[sub]n[/sub] be the greatest integer satisfying a[sub]0[/sub] + a[sub]1[/sub]/10 + ... + a[sub]n[/sub]/10[sup]n[/sup] ≤ x < a[sub]0[/sub] + a[sub]1[/sub]/10 + ... + (a[sub]n[/sub] + 1)/10[sup]n[/sup]. Then 0 ≤ a[sub]n[/sub] ≤ 9 and we have r[sub]n[/sub] ≤ x < r[sub]n[/sub] + 1/10[sup]n[/sup], where r[sub]n[/sub] = a[sub]0[/sub].a[sub]1[/sub]a[sub]2[/sub]...a[sub]n[/sub]. This completes the proof.

It is easy to see from this theorem that we can define an infinite decimal representation of x. For x = 1, we see that a[sub]0[/sub] = 0, and a[sub]n[/sub] = 0 for all n > 1. Then 1 = 1.000....

Since these proofs are virtually identical, you can see that saying 0.999... ≠ 1 is basically the same as saying 1.000... ≠ 1.

Yes, the sphere is the one-point compactification of the plane; this doesn't change the coloring. A map on the surface of a torus may sufficiently be colored with 7 different colors. I can discuss how one determines the number of colors required for a map on a given closed surface, it is related to the genus of the surface (which is why the sphere and the plane have the same sufficient number of colors, but a torus is different), if you wish (this is a generalization of the four color theorem to any closed surface, not just the plane). Interestingly, the Klein bottle is the one exception to the rule: it is sufficiently colored with 6 colors, although in theory this should be 7 colors.

But it is a natural number. Since 0.999... = 1, it is a positive integer and thus a natural number. Just because it doesn't look like it is doesn't mean it isn't. One big reason people have trouble accepting the fact that 0.999... = 1 is that they just don't look the same. But in reality it is just the same as saying 5(3n + 6)/(15n + 30) = 1, they clearly don't look the same, but you'd never say that they aren't equal.

Theorem. Assume x ≥ 0. Then for every integer n ≥ 1 there is a finite decimal r[sub]n[/sub] = a[sub]0[/sub].a[sub]1[/sub]a[sub]2[/sub]...a[sub]n[/sub] such that r[sub]n[/sub] < x ≤ r[sub]n[/sub] + 1/10[sup]n[/sup].

Proof. Let S be the set of all nonnegative integers less than x. Then S is nonempty, since 0 ∈ S, and S is bounded above by x. Then by the Completeness Axiom of the real numbers, S has a supremum, which we denote a[sub]0[/sub] = sup S. a[sub]0[/sub] ∈ S, so that a[sub]0[/sub] is a nonnegative integer. a[sub]0[/sub] is then the greatest integer in x minus one, [x] - 1. Clearly, a[sub]0[/sub] < x ≤ a[sub]0[/sub] + 1. Now let a[sub]1[/sub] = [10x - 10a[sub]0[/sub]] - 1, the greatest integer in 10x - 10a[sub]0[/sub] minus one. Since 0 < 10x - 10a[sub]0[/sub] = 10(x - a[sub]0[/sub]) ≤ 10, we have that 0 ≤ a[sub]1[/sub] ≤ 9 and a[sub]1[/sub] < 10x - 10a[sub]0[/sub] ≤ a[sub]1[/sub] + 1. Then a[sub]1[/sub] is the largest integer satisfying a[sub]0[/sub] + a[sub]1[/sub]/10 < x ≤ a[sub]0[/sub] + (a[sub]1[/sub] + 1)/10. More generally, having chosen a[sub]1[/sub],..., a[sub]n-1[/sub] with 0 ≤ a[sub]i[/sub] ≤ 9, let a[sub]n[/sub] be the greatest integer satisfying a[sub]0[/sub] + a[sub]1[/sub]/10 + ... + a[sub]n[/sub]/10[sup]n[/sup] < x ≤ a[sub]0[/sub] + a[sub]1[/sub]/10 + ... + (a[sub]n[/sub] + 1)/10[sup]n[/sup]. Then 0 ≤ a[sub]n[/sub] ≤ 9 and we have r[sub]n[/sub] < x ≤ r[sub]n[/sub] + 1/10[sup]n[/sup], where r[sub]n[/sub] = a[sub]0[/sub].a[sub]1[/sub]a[sub]2[/sub]...a[sub]n[/sub]. This completes the proof.

It is easy to see from this theorem that we can define an infinite decimal representation of x. For x = 1, we see that a[sub]0[/sub] = 0, and a[sub]n[/sub] = 9 for all n > 1. Then 1 = 0.999....