# Showing that V is a direct sum of two subspaces

Hi guys, I have this general question.

If we are asked to show that the direct sum of ##U+W=V##where ##U## and ##W## are subspaces of ##V=\mathbb{R}^{n}##, would it be possible for us to do so by showing that the generators of the ##U## and ##W## span ##V##? Afterwards we show that their intersection is a zero-vector. For example:

##U## is a subspace generated by ##(0,1)## and ##W## is a subspace generated by ##(2,2)##. Clearly those generators span two dimensional ##V##, and their intersection is ##(0,0)##. Therefore the conclusion can be made that their direct sum is ##V##. Is this kind of reasoning okay?

## Answers and Replies

jbunniii
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Yes, this reasoning is valid. In general, ##V## is the direct sum of ##U## and ##W## if and only if ##V = U+W## and ##U \cap W = \{0\}##. However, this is not the definition of direct sum, it's a (simple) theorem which you should try to prove.

However, be aware that this only works for two subspaces. If you have three or more subspaces, say ##\{U_i\}_{i=1}^{N}##, then it's possible to have ##V = U_1 + \ldots U_N## and ##U_i \cap U_j = \{0\}## for all ##i \neq j##, but the sum is not direct. It's a good exercise to construct an example where this occurs.

Yes, this reasoning is valid. In general, ##V## is the direct sum of ##U## and ##W## if and only if ##V = U+W## and ##U \cap W = \{0\}##. However, this is not the definition of direct sum, it's a (simple) theorem which you should try to prove.

However, be aware that this only works for two subspaces. If you have three or more subspaces, say ##\{U_i\}_{i=1}^{N}##, then it's possible to have ##V = U_1 + \ldots U_N## and ##U_i \cap U_j = \{0\}## for all ##i \neq j##, but the sum is not direct. It's a good exercise to construct an example where this occurs.

Ah ok thanks for your explanation jbuniii. If it's not too much, I would also like to ask why this is true? ##V_1 - V_2 = V_1 + V_2## if we define subtraction to be ##\{ x-y | x \in V_1, y \in V_2 \}.## This doesn't feel intuitive at all, I thought the subtraction of two subspaces will somehow reduce the elements of the new subspace, why it could be equal to the sum of subspaces?

jbunniii
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Ah ok thanks for your explanation jbuniii. If it's not too much, I would also like to ask why this is true? ##V_1 - V_2 = V_1 + V_2## if we define subtraction to be ##\{ x-y | x \in V_1, y \in V_2 \}.## This doesn't feel intuitive at all, I thought the subtraction of two subspaces will somehow reduce the elements of the new subspace, why it could be equal to the sum of subspaces?
Since ##y \in V_2## if and only if ##y \in -V_2##, it doesn't matter whether you add or subtract elements of ##V_2##, you get the same result in both cases: ##V_1 - V_2 = V_1 + V_2##.

Admittedly, the notation ##V_1 - V_2## can seem a bit misleading. Fortunately, since it is the same as ##V_1 + V_2##, there's no reason to use ##V_1 - V_2##.

Note that you could even define ##aV_1 + bV_2 = \{a x + b y | x \in V_1, y \in V_2\}##, where ##a## and ##b## are nonzero scalars. Then ##V_1 - V_2## is just a special case with ##a=1## and ##b=-1##.

Once again we have ##aV_1 + bV_2 = V_1 + V_2## for any nonzero ##a,b##, because ##x \in aV_1## if and only if ##x \in V_1## and similarly, ##y \in V_2## if and only if ##y \in bV_2##. So we can just stick with the notation ##V_1 + V_2## since all the other "linear combinations" of ##V_1## and ##V_2## give the same result.

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jbunniii
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By the way, here's a bit of extra info in case it is helpful. Suppose we start with a collection of subspaces ##\{U_i\}_{i=1}^{N}## of ##V##. We may be interested in combining the ##U_i##'s to form a larger subspace. The naive thing to do would be to form the set theoretic union ##\cup_{i=1}^{N}U_i##. But this won't generally be a subspace.

So what we really want is the smallest subspace containing all of the ##U_i## (or equivalently, the smallest subspace containing ##\cup_{i=1}^{N}U_i##). This turns out to be exactly ##U_1 + \ldots + U_N##. Proof: certainly ##U_1 + \ldots + U_N## is a subspace containing each ##U_i##. If ##S## is another such subspace then it must contain all elements of the form ##u_1+\ldots+u_N## with ##u_i \in U_i##. Thus ##U_1 + \ldots + U_N \subset S## so ##U_1 + \ldots + U_N## is the smallest such subspace.

Now in general, the ##U_i##'s may not be "linearly independent" of each other: maybe there is some nonzero element of ##U_1## which can be expressed as a linear combination of elements of the other ##U_i##'s. But if they ARE linearly independent, then we say that the sum is a direct sum, and we write it as ##U_1 \oplus \ldots \oplus U_N## instead of ##U_1 + \ldots + U_N##. We can obtain a basis for a direct sum ##U_1 \oplus \ldots \oplus U_N## by simply selecting a basis ##B_i## for each ##U_i## and taking the union: ##B = \cup_{i=1}^{N} B_i##. Indeed, this union is disjoint because there are no common elements among the ##B_i##'s. In the finite-dimensional case, this immediately tells us that
$$\text{dim}(U_1 \oplus \ldots \oplus U_N) = \sum_{i=1}^{N}\text{dim}(U_i)$$
This is not true if the sum is not direct. In that case, all we can say is that
$$\text{dim}(U_1 + \ldots + U_N) \leq \sum_{i=1}^{N}\text{dim}(U_i)$$

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Wow thanks for your detailed explanation. I just want to make sure for this last time that I understand this correctly.

Since ##y \in V_2## if and only if ##y \in -V_2##, it doesn't matter whether you add or subtract elements of ##V_2##, you get the same result in both cases: ##V_1 - V_2 = V_1 + V_2##.

Can I also intuitively think that ##V_2## = ##-V_2## if ##V## is a subspace, because if you multiply every elements in the subspace with -1, then you will basically get the same subspace because every element has an additive inverse? (In other words the negatives will become the positives and vice-versa)

jbunniii
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Can I also intuitively think that ##V_2## = ##-V_2## if ##V## is a subspace, because if you multiply every elements in the subspace with -1, then you will basically get the same subspace because every element has an additive inverse? (In other words the negatives will become the positives and vice-versa)
Sure, the fact that ##V_2## is a subspace and therefore contains its additive inverses is exactly the reason why ##y \in V_2## if and only if ##y \in -V_2##, and this latter statement is the definition of set equality: we conclude that ##V_2 = -V_2##.


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# Verify that $\binom{n+1}{4} = \frac{\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)}{3}$ for $n \geq 4$

Verify that for $n \geq 4$ $$\dbinom{n+1}{4} = \frac{\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)}{3}$$

Now present a combinatoric argument for the above.

First, by verify does it mean check for some n > 3? if so, then n = 4 gives both sides value 5. I have tried expanding both sides and It just gets messy, but i'm sure that would be attempting to prove it?

I cant quite move ahead with this one, I have said that $\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)$ is the number of ways of choosing 2 pairs of objects from n objects. my reasoning for this is that $\dbinom{n}{2}$ is the number of ways of choosing 2 objects from n and hence $\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)$ is the number of ways of choosing 2 of these ways... What i am struggling to do is understand how the three comes into it.

• I think, in this case, "verify" means to prove algebraically, by manipulating the right hand side until it looks like the left. – Theo Bendit Jul 17 '15 at 18:08
• Please don't use display math in titles; it takes too much vertical space on the front page. – Rahul Jul 17 '15 at 18:09
• Hints: (1) How are two pairs of two things like one quadruple of things? (2) What if both pairs contain a common element? These are two independent hints. – Rahul Jul 17 '15 at 18:12
• before reading the answer below fully and do the bad thing, i would like to take your comment into consideration and work it out for myself. (1) A pair of things is like one quadruple of things when each pair are disjoint, two pairs of things are not like a quadruple of things when they have a common member. so to expand, if i have four objects a,b,c,d. there are three ways i can choose distinct pairs from these four objects. [ab][cd], [bc][ad], [ac][bd]. Now concerning (2), i am struggling to use this hint at the moment. – user197848 Jul 17 '15 at 18:47

"Verify" would mean verify algebraically (using formulas such as $\binom{n+1}{4}=\frac{(n+1)n(n-1)(n-2)}{24}$. It's not that hard if you just consider the factorizations of each side. However, there is a nice combinatorial argument. $\left(\substack{{\binom{n}{2}}\\{\displaystyle 2}}\right)$ is the number of ways to pick two pairs of elements from a set of $n$; we can combine these pairs, and if an element is common to both pairs, then treat one of the instances of the element as a new $(n+1)$-st element. This counts the number of ways to pick 4 elements from a set of $n+1$. However, note that all ways to pick 4 are counted exactly 3 times, so you must divide by 3 to get the equality that is desired.

To verify:

$${n\choose2} = \frac{n(n-1)}{2}$$

$${\frac{n(n-1)}{2}\choose2} = \frac{\frac{n(n-1)}{2}.(\frac{n(n-1)}{2}-1)}{2}$$

$${\frac{n(n-1)}{2}\choose2} = \frac{n(n-1).(n^2-n-2)}{8} = \frac{(n+1)n(n-1)(n-2)}{8}\tag1 - RHS$$

$${(n+1)\choose4} = \frac{(n+1)n(n-1)(n-2)}{4!} = \frac{RHS}{3}$$

Hence proved.

• AH i see where it went all wrong for me. thanks :) – user197848 Jul 17 '15 at 18:20
• You are welcome. See the other responder's answer for a combintorial argument. It makes sense. Good luck – Satish Ramanathan Jul 17 '15 at 18:22

Using \begin{align} \binom{n}{2} = \frac{n(n-1)}{2} \end{align} then \begin{align} \frac{1}{3} \, \binom{\binom{n}{2}}{2} &= \frac{1}{2} \binom{n}{2} \, \left(\frac{n}{2} - 1\right) \\ &= \frac{n(n-1)}{4!} \left( n(n-1)-2 \right) = \frac{(n+1)(n)(n-1)(n-2)}{4!} \\ &= \binom{n+1}{4}. \end{align}

As to the modified component of the question: The comments seem to be helpful

Algebraic Manipulation is actually not that bad. $\binom{n}{2}=n(n-1)/2$, so $$\binom{\binom{n}{2}}{2}=\binom{n(n-1)/2}{2}=(n(n-1)/2)(n(n-1)/2-1)/6=n(n-1)(n^2-n-2)/24=n(n-1)(n+1)(n-2)/24=(n+1)!/4!(n-3)!=\binom{n+1}{4}$$


# Help finding the limit of this series $\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots$

How can I go about finding the limit of $$\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots = \sum_{k = 1}^{\infty} \frac{1}{2^{k+1}}?$$ Could I use the absolute value theorem? I have a feeling it converges to $0$ but I am not sure.

• Of the series or the sequence? – Sloan Jul 2 '15 at 17:26
• sorry for the confusion. The third one and the limit of the series. – Lil Jul 2 '15 at 17:30
• Divide every term by $2$ and see what happens. – Yves Daoust Jul 3 '15 at 12:28
• The sequence converges to zero, but the sum of the terms converges to a positive value. Be sure you make a distinction between the values of the individual terms and the sum of the terms. – John Molokach Jul 3 '15 at 16:12

You know the limit of $1+\frac12 +\frac14+\frac18+\ldots$, don't you?

Your sequence is just like that, but without the $1$ and the $\frac12$.

Using the formula for the geometric series $$\sum\limits_{k=0}^\infty q^k=\frac{1}{1-q}$$ with $|q|<1$ we have: $$\sum\limits_{k=0}^\infty \frac{1}{2^{n+1}}=\sum\limits_{k=0}^\infty \left(\frac{1}{2}\right)^{n+1}=\sum\limits_{k=0}^\infty \frac{1}{2}\cdot \left(\frac{1}{2}\right)^n = \frac {1}{2}\cdot\sum\limits_{k=0}^\infty \left(\frac{1}{2}\right)^n = \frac {1}{2}\cdot \frac{1}{1-\frac{1}{2}}=\frac{1}{2}\cdot 2=1.$$

• The problem was modified after you answered it. – N. F. Taussig Jul 3 '15 at 10:03
• Well, if we start with $k=1$ we can easily obtain $$\sum\limits_{k=1}^\infty \frac {1}{2^{n+1}}=\sum\limits_{k=0}^\infty \frac{1}{2^{n+1}}-\frac{1}{2^{0+1}}=1-\frac{1}{2}=\frac {1}{2}.$$ – Hirshy Jul 3 '15 at 10:17

If your series begins at $\frac{1}{4}$, then it should be:

The sequence $a_n = \dfrac{1}{2^{n+2}}$ converges to $0$ as $n \to \infty$.

The series $$\sum_{n=0}^{\infty} \frac{1}{2^{n+2}}=\frac{1}{4} \cdot \sum_{n=0}^{\infty} \frac{1}{2^{n}} = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots$$ is the sum of a geometric sequence with common ratio $\frac{1}{2}$ and first term $\frac{1}{4}$. The series is convergent since the absolute value of the common ratio, $|r| = \frac{1}{2} < 1$. The sum evaluates to $$\frac{\frac{1}{4}}{1- \frac{1}{2}} = \frac{1}{2}$$ by using the general formula for the sum of a geometric series with common ratio $r$ and first term $a$, namely:$$\sum_{n=0}^{\infty}ar^n = \frac{a}{1-r}$$ as long as $|r| < 1$.

• maybe I have the wrong formula.. my series is 1/4+1/8+1/16+1/32... – Lil Jul 2 '15 at 17:54
• It begins at 1/4 not 1/2 as everyone is commenting. – Lil Jul 2 '15 at 17:54
• Then your series is $$\sum_{n=0}^{\infty} \frac{1}{2^{n+2}}$$ since the sum of a geometric series starts from $n=0$ by convention. :-) – Zain Patel Jul 2 '15 at 17:57
• I don't understand why it would be 1/2^n+2 because when n=1 it would become 1/2^3 which is 1/8 but my first term is 1/4...? – Lil Jul 2 '15 at 17:58
• doesn't the sum of a geometric series begin at 1? or no? – Lil Jul 2 '15 at 17:58

Use $$\sum_{n=0}^\infty a r^n=\frac{a}{1-r},\quad |r|<1.$$ Take $r=\frac12$ and $a=\frac14$ since the first term of your series is $\frac14$ and the common ratio of your Geometric progression is $\frac12.$

• The problem was modified after you answered it. – N. F. Taussig Jul 3 '15 at 10:03

I will add an indefinite version of the sum. In difference calculus we have that the function $2^n$ is the analogous of $e^x$ on ordinary differential calculus.

We have that $\sum 2^n \delta n=2^n$ (you can see it by yourself if you do the difference and see that it doesnt change, i.e. $\Delta 2^n=2^{n+1}-2^n=2^n$).

In your sum, doesnt taking limits, we have that

$$\sum 2^{-n-1}\delta n=-2^{-n}$$

If we take limits then

$$\sum_{n=1}^{\infty} 2^{-n-1}=-2^{-n}\bigg\lvert_{1}^{\infty}=-2^{-\infty}+2^{-1}=\frac{1}{2}$$

Take care with notation because

$$\sum_{k\ge 0}2^k\ne\sum_{k\ge 0}2^n$$


# Every normal subgroup is the kernel of some homomorphism

Clearly the kernel of a group homomorphism is normal, but I often hear my professor mention that any normal subgroup is the kernel of some homomorphism.

This feels correct but isn't entirely obvious to me.

One thought I had is that for any normal subgroup $N$ of $G$, we could define the quotient homomorphism $\pi:G\to G/N$ since $G/N$ is a group.

I was imagining that we could consider $\pi^{-1}:G/N\to G$, whose kernel would then be $N$. However, $\pi^{-1}$ doesn't exist since $\pi$ is not a bijection in general.

So my question is this: is there an obvious way to define a homomorphism whose kernel is an arbitrary normal subgroup of $G$? Or does it depend on the particular group whether you can define such a homomorphism?

• No, you more or less have it, the kernel of $\pi\colon G\to G/N$ is $N$. No need to worry about $\pi^{-1}$, or if it exists. Jul 18, 2015 at 23:17

You only have to consider that $\pi:G\to G/N$, defined by $\pi(x)=xN$, is a homomorphism and the $\ker \pi$ is precisely $N$.

• overcomplicating as usual. thanks. Jul 18, 2015 at 23:18
• that's why we are meant to be here Jul 18, 2015 at 23:19

Anyway, if the kernel is non-trivial, the homomorphism is not injective!

You're confusing the inverse of an isomorphism, and the inverse image of a subset by a map. So, yes, if $$N$$ is a normal subgroup, it is the kernel of the canonical homomorphism : \begin{align*}\pi\colon G&\longrightarrow G/N,\\ g&\longmapsto gN. \end{align*} Indeed, $$\pi^{-1}(\bar 1)=\pi^{-1}(N)=N$$.

Your intuition is correct, and unfortunately, yes, $\pi^{-1}$ isn't a function. Could we somehow "make" it one?

Well, one way this is often done with "ordinary" functions, is to treat $f^{-1}$, for a given function $f: A \to B$, as not something like this:

$f^{-1}: B \to A$

but instead considering $f^{-1}$ as a function between power sets:

$f^{-1}: \mathcal{P}(B) \to \mathcal{P}(A)$

where for a subset $Y \subseteq B$, we define $f^{-1}(Y) = X = \{x \in A: f(x) \in Y\}$.

Applying this definition to $\pi^{-1}$, we get:

$\pi^{-1}(e_{G/N}) = \pi^{-1}(N) = \{g \in G: \pi(g) \in N\}$

$= \{g \in G: gN = N\} = \{g \in G: ge^{-1} \in N\} = \{g \in G: g \in N\} = N$


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## AP®/College Statistics

### Course: AP®/College Statistics>Unit 4

Lesson 4: Normal distributions and the empirical rule

# Basic normal calculations

AP.STATS:
VAR‑2 (EU)
,
VAR‑2.A (LO)
,
VAR‑2.A.3 (EK)
CCSS.Math:
Many measurements fit a special distribution called the normal distribution. In a normal distribution,
• approximately equals, 68, percent of the data falls within 1 standard deviation of the mean
• approximately equals, 95, percent of the data falls within 2 standard deviations of the mean
• approximately equals, 99, point, 7, percent of the data falls within 3 standard deviations of the mean
Problem 1
A large sample of females had their systolic blood pressure measured. The mean blood pressure was 125 millimeters of mercury and the standard deviation was 10 millimeters of mercury.
Which normal distribution below best summarizes the data?

Problem 2
What percent of females had blood pressures between 105 and 135 millimeters of mercury?
percent

Problem 3
The sample had a total of 400 females that participated.
About how many females in the sample had blood pressures higher than 145 millimeters of mercury?
approximately equals
females

## Want to join the conversation?

• What is the Gauss curve have to do with this?
• For the empirical rule to be valid the data must be normally distributed, so the rules for percentages in the problems above would not hold true if the data didn't follow a gaussian or normal distribution.
• I answered the last question in the following way. I figured out the z score which was 0.9772 for value that is two standard deviations above normal. Therefore one can deduce that the sample size that lies to the right of this should be 0.0228 (1-0.9772). That multiplied by 400 gives me 9.1 approximated to 9. What is wrong with this method?
• I agree that using the z-Score is the most accurate answer. The answer given is using approximations of the normal and are not as accurate.
• where does the 13.5% come from in question 3?
• Empirical rule!
Ok, so 95% of the observations are within 2 standard deviations of the mean, and 68% of the observations are within 1 standard deviation of the mean, right?
Let's find the difference between 2 s.d. and 1 s.d. It will be 95%-68%=27%. But you have to divide this 27% by 2 because you have to find the percentage between 105 and 115 millimeters. There comes 13.5%!
• why do i have E.D.?
• what if they ask you to solve for standard deviation with only knowing the range
• This question is not really meaningful for a normal distribution, since all normal distributions have infinite range.

For general data sets, knowing the range of a data set is not sufficient for finding its standard deviation. For example, the data sets 1,5,5,9 and 1,2,8,9 both have range 9-1=8, but 1,2,8,9 has the larger standard deviation because the values are spread out farther from the mean (5).

Have a blessed, wonderful day!
• Are these realistic statistics? They better not be, my math teacher commonly uses extremely unrealistic stats, it's funny.
• For the titles of Articles, Videos, and Skills; should the wording follow the same grammar rules as the title of a book does?
e.g.: Should the titles be capitalized with the exception of conjunctions like and/or/but/ect?

I'm slightly confused, because KA only capitalizes the first word, but I can't find anywhere on the web where it tells you how to properly right the title.

Any help?

I'm slightly confused...
• Khan Academy is an exception to the usually title rules. Sometimes websites might only capitalize the first word or not follow the exact rules as in a book for some videos and articles. You are correct though, that rule is used commonly in books, movies, etc.
(1 vote)
• Is there a table or cheat sheet I can use to help with the subject?
(1 vote)
• Here is a website that has z-tables for positive and negative z-scores as well as other things related to this subject: https://z-scoretable.com/.

There are plenty of z-tables on Google Images as well.

Hope this helps!😄


## Wednesday, December 31, 2014

### Almost all integers contain the digit 3

This is my last chance to blog this year. However, I did not get back into the proof of Gödel's incompleteness theorem yet. So instead, this post is motivated by this Numberphile video, which explains why almost all integers contain the digit 3. More precisely, the percentage of integers that contain at least one occurrence of the digit '3' in the set $$\mathbb{N}_u = \{0,1,2,3,\cdots,u-1\}$$ goes to 100% as the upper bound $$u$$ goes to infinity.

We are going to attack this problem with two different approaches. First, we'll use a recurrence relation approach. Second, we'll go over the combinatorial proof discussed in the video. Finally, we'll check that the two approaches do give the same answer.

In both approaches, we will use values of $$u$$ that are powers of 10. So if $$u = 10^n$$, for $$n \in \mathbb{N}^+$$, then the set under consideration $$\mathbb{N}_{10^n}$$, which we denote $$A_n$$, is the set of all of the (non-negative) integers containing at most $$n$$ digits.

Recurrence relation approach

In this approach, we use $$S_n$$, for $$n \in \mathbb{N}^+$$, to denote the subset of $$A_n$$ of all of the integers that contain at least one occurrence of the digit '3' and we use $$T_n$$ to denote $$|S_n|$$, the cardinality of $$S_n$$.
• Since $$A_1=\{0,1,2,3,4,5,6,7,8,9\}$$, $$S_1=\{3\}$$ and $$T_1=1$$.
• Since $$A_2=\{0,1,2,3,4,\cdots,98,99\}$$, $$S_2=\{ 3,13,23,30,31,32,33,34,\cdots,39,43,53,63,73,83,93\}$$$$=\{ 3,13,23,43,53,63,73,83,93\}$$$$\cup \{30,31,32,33,34,\cdots,39\}$$. In other words, we can split the set $$S_2$$ into two subsets, one whose elements all start with the digit '3'  (there are exactly 10 such integers), and the other one whose elements do not start with the digit '3' (there are exactly $$9\cdot T_1$$ of those, since if the first digit is not '3' then the least significant digit must belong to $$S_1$$). Thus, $$T_2 = 10 + 9\cdot T_1 = 10 + 9 = 19$$.
• Similarly, since $$A_3=\{0,1,2,3,4,\cdots,998,999\}$$, $$S_3$$ is the union of two sets,  one whose elements all start with the digit '3'  (there are exactly $$10^2$$ such integers), and the other one whose elements do not start with the digit '3' (there are exactly $$9\cdot T_2$$ of those, since if the most significant digit is not '3' then the number formed by the other digits must belong to $$S_2$$). Thus, $$T_3 = 10^2 + 9\cdot T_2= 100 + 9\cdot 19 = 271$$.
• Since this reasoning clearly generalizes to all values of $$n \in \mathbb{N}^+$$, we obtain the following recurrence relation:
$\left\{ \begin{array}{ll} T_1 = 1 & \\ T_n = 10^{n-1} + 9T_{n-1} & \text{if}\ n>1 \end{array} \right.$ Now, let's solve this recurrence relation by iteration to get a closed-form formula for $$T_n$$:

$$\begin{array}{l@{0pt}l} T_1 & = 1\\ T_2 = 10^1 + 9T_1 & = 10+9 \\ T_3 = 10^2 + 9T_2 & = 10^2+9(10+9) \\ & = 10^2 + 9\cdot 10 + 9^2 \\ T_4 = 10^3 + 9T_3 & = 10^3+9(10^2 + 9\cdot 10 + 9^2) \\ & = 10^3+9\cdot 10^2 + 9^2 \cdot 10 + 9^3\\ T_5 = 10^4 + 9T_4 & = 10^4+9(10^3+9\cdot 10^2 + 9^2 \cdot 10 + 9^3)\\ & = 10^4+9\cdot 10^3+9^2\cdot 10^2 + 9^3 \cdot 10 + 9^4\\ \cdots &\\ T_n = 10^{n-1} + 9T_{n-1} & = 10^{n-1}+9\cdot 10^{n-2}+9^2\cdot 10^{n-3} + \cdots + 9^{n-2}\cdot 10^1 + 9^{n-1}\\ \end{array}$$

In conclusion, $$\displaystyle T_n = \sum_{i=0}^{n-1} \left( 9^{n-i-1}\cdot 10^i\right)$$ for $$n \in \mathbb{N}^+$$

Since this formula is rather ugly, let's turn to the combinatorial approach discussed in the Numberphile video.

Combinatorial approach

It is easy to determine the total number of integers with exactly $$n$$ digits without having to enumerate them, namely $$10 \times 10 \times \cdots \times 10 = 10^n$$,  since there are exactly 10 digits to choose from for each position in the $$n$$-digit number. Note that this count actually includes all of the numbers with leading zeros, such as 01 and 00023, which are identical to 1 and 23, respectively. In other words, what we really have is  $$\left|A_n\right|= 10^n$$.

Similarly, we can compute the total number of integers with at most $$n$$ digits that do not contain the digit '3', namely $$9 \times 9 \times \cdots \times 9 = 9^n$$, since there are now only 9 digits to choose from at each position in the integer.

In conclusion, $$T_n = 10^n -9^n$$.

This is both a much nicer and easier-to-derive closed-form formula than the one we obtained with the recurrence relation approach. But are the two formulas equal?

Let's check our work

Let $$P(n)$$, for $$n \in \mathbb{N}^+$$, denote: $$\displaystyle \sum_{i=0}^{n-1} \left( 9^{n-i-1}\cdot 10^i\right) = 10^n - 9^n$$

We now prove by mathematical induction that $$P(n)$$ holds for  $$n \in \mathbb{N}^+.$$

Basis:
•  $$\displaystyle \sum_{i=0}^{1-1} \left( 9^{1-i-1}\cdot 10^i\right) = \sum_{i=0}^{0} \left( 9^{1-i-1}\cdot 10^i\right) = 9^{1-0-1}\cdot 10^0 = 9^0 \cdot 10^0 =1$$
• $$10^1 - 9^1 = 10 - 9 = 1$$
• Therefore $$P(1)$$ holds
Inductive step:
• Assume that $$P(n)$$ holds for any $$n \in \mathbb{N}^+$$, that is: $\sum_{i=0}^{n-1} \left( 9^{n-i-1}\cdot 10^i\right) = 10^n - 9^n \quad \text{[inductive hypothesis]}$
• Let's compute the left-hand side of $$P(n+1)$$:
$\begin{array}{l@{0pt}l} \sum_{i=0}^{n} \left( 9^{n-i}\cdot 10^i\right) & = 9\left(\frac{1}{9}\sum_{i=0}^{n} \left( 9^{n-i}\cdot 10^i\right)\right)\\ & = 9\left(\sum_{i=0}^{n} \left( 9^{n-i-1}\cdot 10^i\right)\right)\\ & = 9\left(\sum_{i=0}^{n-1} \left( 9^{n-i-1}\cdot 10^i\right) + \sum_{i=n}^{n} \left( 9^{n-i-1}\cdot 10^i\right) \right)\\ & = 9\left(\sum_{i=0}^{n-1} \left( 9^{n-i-1}\cdot 10^i\right) + \left( 9^{n-n-1}\cdot 10^n\right) \right)\\ & = 9\left(\sum_{i=0}^{n-1} \left( 9^{n-i-1}\cdot 10^i\right) + \frac{10^n}{9} \right)\\ & = 9\left( \left(10^n - 9^n\right) + \frac{10^n}{9} \right) \quad \text{by the inductive hypothesis} \\ & = 9\cdot 10^n - 9\cdot 9^n + 10^n\\ & = 10\cdot 10^n - 9\cdot 9^n\\ & = 10^{n+1} - 9^{n+1}\\ \end{array}$
• Therefore $$P(n+1)$$ holds.

Main result

We are now in a position to prove our main result, namely that the percentage of integers that contain the digit '3' in the set $$A_n$$  goes to 100% as $$n$$ goes to infinity. This percentage is equal to
$$100\times\frac{T_n}{\left|A_n\right|}=100\times\left(\frac{10^n-9^n}{10^n}\right)=100\times\left(1-\frac{9^n}{10^n}\right)= 100 - 100\left(\frac{9}{10}\right)^n$$.
And this percentage goes to 100% as $$n$$ goes to infinity, since $$\displaystyle \lim_{n \to \infty}\left( \frac{9}{10}\right)^n = 0$$.

Discussion

Of course, all of the proofs above still hold if we had picked the digit '8' (say) instead of the digit '3'. In other words, it is also true that almost all integers contain the digit '8'. So, if we use $$E_n$$ to denote the cardinality of the subset of $$A_n$$ of all of the integers that contain at least one occurrence of the digit '8', $$E_n = 10^n-9^n$$. Similarly for the digit '5' (say): $$F_n = 10^n-9^n$$.

But, if both $$100\times\frac{T_n}{\left|A_n\right|}$$ and  $$100\times\frac{E_n}{\left|A_n\right|}$$ go to 100% as $$n$$ goes to infinity, the sum of these two percentages will eventually exceed 100%! That is true. However, this sum double counts all of the integers that contain both the digit '3' and the digit '8'. The correct percentage of such integers is $$100\times\frac{10^n-8^n}{10^n}$$, which also goes to 100% as $$n$$ goes to infinity.

So there is no paradox here. When $$n$$ gets large enough, almost all of the integers will contain all of the digits 0 through 9. Note that all of these results are asymptotic. Indeed, it takes relatively large values of $$n$$ for these percentages to get anywhere close to 100%. For example, for the percentage of integers that contain the digit '3' to reach 90%, 95%, 99% and 99.99%, the number $$n$$ of digits in the integer must be larger than 21, 28, 43 and 87, respectively.

## Tuesday, July 9, 2013

### Robinson Arithmetic is Σ1-complete

Recall that Q (i.e., Robinson Arithmetic) is an axiomatized formal theory (AFT) of arithmetic couched in the interpreted formal language LA. Let L be a subset of LA and let T be some AFT of arithmetic.

We say that T is L-sound iff, for any sentence φ in L, if T ⊢ φ, then φ is true.

We say that T is L-complete iff, for any sentence φ in L, if φ is true, then T ⊢ φ.

In chapter 9, Peter Smith defines a subset of LA, called Σ1. Then, using the fact that Q is order-adequate, he proves that Q is Σ1-complete. This is important because the well-formed formulas (wff's) of Σcan express the decidable numerical properties and relations, and therefore Q will be sufficiently strong. Now to the details...

First, let's define a few interesting subsets of LA:

## Sunday, July 7, 2013

In chapter 9, Peter Smith defines the following concept: A theory T that captures the relation ≤ is order-adequate if it satisfies the following nine properties:
• O1: T ⊢ ∀x (0 ≤ x)
• O2: For any n, T ⊢ ∀x ((x = 0 ∨ x = 1 ∨ x = 2 ∨ ... ∨ x = n) → x ≤ n)
• O3: For any n, T ⊢ ∀x (x ≤ n → (x = 0 ∨ x = 1 ∨ x = 2 ∨ ... ∨ x = n))
• O4: For any n, if T ⊢ φ(0) and T ⊢ φ(1) and ... and T ⊢ φ(n) then T ⊢ (∀x ≤ n)φ(x)
• O5: For any n, if T ⊢ φ(0) or T ⊢ φ(1) or ... or T ⊢ φ(n) then T ⊢ (∃x ≤ n)φ(x)
• O6: For any n, T ⊢ ∀x (x ≤ n → x ≤ Sn)
• O7: For any n, T ⊢ ∀x (n ≤ x  → (n = x  ∨ Sn ≤ x))
• O8: For any n, T ⊢ ∀x (x ≤ n ∨ n ≤ x)
• O9: For any n>0, T ⊢ (∀x ≤ n-1)φ(x) → (∀x ≤ n)(x ≠ n → φ(x))
Then, we have the following theorem:

## Wednesday, July 3, 2013

### Robinson Arithmetic captures the "less-than-or-equal-to" relation

In this post, we'll start discussing the material in Chapter 9 of Peter Smith's book, namely up to section 9.3.

Before proceeding, review the definition of Robinson Arithmetic (denoted Q) as well as what it means for a theory to capture a numerical relation.

Now, we'll show that Robinson Arithmetic captures the ≤ numerical relation with the open well-formed formula: ∃x (x + m = n).

In other words, we are going to prove the following two-part theorem:
Theorem: If m and n are natural numbers, then:
1. If m ≤ n, then Q ⊢ ∃x (x + m = n)
2. If m > n, then Q ⊢ ¬ ∃x (x + m = n)
Recall that if m and n are natural numbers, then m and n are the numerical terms representing m and n, respectively, in the formal theory.

Proof sketch of part 1:

## Sunday, June 30, 2013

### Q - Robinson Arithmetic

Now that we are familiar with Baby Arithmetic (BA), we can make its language more expressive by allowing variables and quantifiers back into its logical vocabulary. When we do this, we simply obtain the interpreted language LA, that was described earlier.

Since we now have variables and quantifiers, we can replace the schemata of BA with regular axioms (see below). The resulting formal system of arithmetic is called Robinson Arithmetic and is often denoted by the letter Q, as described in Chapter 8 of Peter Smith's book. Here is his definition of Q:

## Friday, June 28, 2013

### Baby arithmetic

Since chapter 7 in Peter Smith's book is so short, I'll summarize it quickly and then start discussing chapter 8.

Chapter 7 starts by comparing the two incompleteness theorems discussed in previous chapters (let's call these "Smith's theorems") to Gödel's first incompleteness theorem as follows:

## Thursday, June 27, 2013

### Consistent, sufficiently strong formal systems of arithmetic are incomplete

In Chapter 6, Peter Smith proves another incompleteness theorem. To place this theorem in context, let's review some definitions about axiomatized formal theories (or AFTs) from earlier posts.

Let T be some AFT.
• T is consistent iff (if and only if) there is no sentence φ such that T proves both φ and ¬ φ.
• T is sound iff every theorem that T proves is true according to the interpretation that is built into T's language.
• T is (negation-)complete iff for every sentence φ in T's language, T proves either φ or ¬ φ.
• T is decidable iff there is an algorithm that, given any sentence φ in T's language, determines whether or not T proves φ.
• If needed, go here to review what it means for T's language to express a numerical property P or a numerical relation R.
• If needed, go here to review what it means for T to capture a numerical property P or a numerical relation R.
• If needed, go here to review what it means for T's language to be sufficiently expressive.

We'll use the following abbreviations:

## Monday, June 24, 2013

### Sound formal systems with sufficiently expressive languages are incomplete

In chapter 5, Peter Smith uses a counting argument to prove that sound axiomatized formal theories (AFTs) that can express a good amount of arithmetic are incomplete. In short: since their set of theorems is effectively enumerable but their set of true sentences is not, the two sets must be different. In a sound theory, the theorems are all true. Therefore, some truths must remained unproved in such theories.

Let's start with a definition. The language of an AFT of arithmetic is sufficiently expressive if and only if:
1. It can express quantification over numbers, and
2. It can express every effectively decidable two-place numerical relation.
Now, to the pivotal theorem of this chapter:

## Sunday, June 23, 2013

### An enumerable but not effectively enumerable set

Let's focus on the first half of chapter 5 in Peter Smith's book. First, Smith describes a new characterization of effectively enumerable sets. Second, Smith gives an example of a set of integers that is not effectively enumerable.

Let's define the numerical domain of an algorithm as the set of natural numbers with the following property: when the algorithm takes one of these numbers as input, it terminates and returns a result. Every algorithm has a numerical domain. If the input to some algorithm is not a single natural number, then the numerical domain of this algorithm is the empty set. Otherwise, the numerical domain is the set of those natural numbers on which the algorithm does not crash and does not go into an infinite loop.

It turns out that each numerical domain is effectively enumerable. In fact, the converse is also true, according to the following theorem:

## Saturday, June 22, 2013

### Capturing numerical properties in a formal language of arithmetic

Peter Smith starts Chapter 4 by describing LA, a formal language that is at the core of several AFTs (axiomatized formal theories) of arithmetic. Then Smith explains what it means for a formal language of arithmetic to "express" a numerical property and the stronger notion of "capturing" a numerical property.

Here is the definition of LA:

## Friday, June 21, 2013

### Formal systems or axiomatized formal theories

In chapter 3, Peter Smith defines formal systems or, as he calls them, axiomatized formal theories (I will use AFT as an abbreviation for this phrase).

A theory T is an AFT if...

## Wednesday, June 19, 2013

### Effective computability, decidability and enumerability

In Chapter 2, Smith covers familiar ground. So this post will be short. Here is a quick summary, section by section.

Section 1 reviews terminology pertaining to functions, namely the notions of domain and range, as well as special cases of functions, namely injective (or one-to-one) functions, surjective (or onto) functions and bijective functions (or one-to-one correspondences).

## Tuesday, June 18, 2013

### Peter Smith's overview of Gödel's incompleteness theorems

Finally! As was my initial goal, I now turn my attention to Peter Smith's book on the proof of Gödel's incompleteness theorems (by the way, I own the first edition of his book; so that is the one I will be referring to). I did not expect to spend that much time on Nagel and Newman's book. But it was worthwhile in getting the big picture.

Chapter 1 is an introductory and user-friendly discussion of topics that were mentioned in earlier posts, such as basic arithmetic, formal systems, (in)completeness, consistency, the statement of Gödel's incompleteness theorems and some of their implications. Therefore, in this post, I will just highlight the points where Smith provides new information or has a different perspective.

## Monday, June 17, 2013

### Proof sketch of Gödel's incompleteness theorems

In this post, I will follow the outline of the proof given in Section VII of Nagel and Newman's book. Recall that:
• a formal system F is consistent if there is no well-formed formula (wff) w such that F proves both w and ¬ w, and
• a formal system is (semantically) complete if it can prove all of the true wff's that it can express.
Now, Gödel's first incompleteness theorem can be paraphrased as:
Any consistent formal system that is expressive enough to model arithmetic is both incomplete and "incompletable."
and Gödel's second incompleteness theorem can be paraphrased as:
Any consistent formal system that is expressive enough to model arithmetic cannot prove its own consistency.
According to Nagel and Newman, a proof of the first theorem can be sketched in 4 steps:

## Friday, June 7, 2013

### Mappings and the arithmetization of meta-mathematics

In an earlier post, we discussed mappings in general and then described Gödel numbering, which is a mapping between the set of well-formed formulas (wff's) in a formal system and the set of positive integers.

In part B of section VII, Nagel and Newman describe a more interesting mapping used in Gödel's proof. In this mapping, the domain is the set of meta-mathematical statements about structural properties of wff's, while the co-domain is the set of wff's. This post describes how this mapping enabled Gödel to arithmetize meta-mathematics.

## Thursday, June 6, 2013

### Mappings and Gödel numbering

Section VI of Nagel and Newman's book describes mappings. A mapping is an operation that applies to two sets called the domain and the co-domain. More specifically, a mapping associates to each element of the domain one or more elements of the co-domain.

## Tuesday, May 28, 2013

### Absolute proof of consistency of FSN

In section IV of their great little book, Nagel and Newman discuss efforts by Gottlob Frege and then Bertrand Russell to reduce arithmetic to logic. This is clearly another attempt at a relative proof of consistency: if this reduction were successful, then arithmetic would be consistent provided logic is consistent.

Whether this latter statement is true or not, Whitehead and Russell's Principia Mathematica was a landmark achievement: it almost completed the first step in an absolute proof of consistency of arithmetic, since it led to the formalization of an axiomatic system for arithmetic.

In section V, Nagel and Newman describe a formalization of propositional (or sentential) logic, that is, a subset of the logic system in Principia Mathematica (but not a large enough subset to represent arithmetic). The bulk of this section first describes the formalization process, which yields the standard syntax and inference rules of propositional logic (including modus ponens) and then outlines an absolute proof of consistency of this formalized axiomatic system.

This absolute proof of consistency is a proof by contrapositive, which relies on the following true conditional statement:

## Sunday, May 26, 2013

### Absolute proofs of consistency and meta-mathematics

In earlier posts, we explained why consistency is an important property of axiomatic systems and discussed relative proofs of consistency, in which the proof of consistency of a system is based on the assumption that another axiomatic system is consistent. In this post, we introduce absolute proofs of consistency that do not make any assumptions about any other axiomatic system. Apparently, David Hilbert was the first to study and propose such proofs, according to Nagel and Newman's book (Section III, page 26).

Recall that an axiomatic system is consistent if it cannot derive both a theorem and its negation. What do we mean by the negation of a theorem? Let's take, as a simple example, the theorem: "6 is divisible by 3." Its negation is simply the following statement: "6 is not divisible by 3" or equivalently "It is not the case that 6 is divisible by 3." This second formulation of the negation, although less elegant in English, is preferable because the negation is added to the front of the original theorem. In a formal system, negation is handled by simply adding a symbol for the phrase "it is not the case that." Several symbols have been used for negation, such as ~ and ¬ . We'll use the latter here. So, if T is any theorem in some formal system, then the formula  ¬T is the negation of T.

Remember that our formal system FS did not have a symbol for negation. So we will extend FS into a new formal system called FSN (for FS with Negation), whose alphabet is { E, 0, 1, ¬ }. FSN has exactly one axiom, namely the same as A1 in FSFSN also has the same inference rules as FS, namely IR1 and IR2. But it has one additional inference rule that uses negation. Here is the full description of FSN:

## Wednesday, May 22, 2013

### Relative proofs of consistency

In the last post, we defined and explained the importance of the property of consistency for an axiomatic system. The second half of Section II in Nagel and Newman's book describes ways of proving that an axiomatic system is consistent.

But first, note that the question of consistency of Euclidean geometry did not arise. Its axioms were supposed to describe the real world; and something that actually exists cannot be self-contradictory. In other words, existence (or truth) implies internal consistency.

The need for consistency proofs arose much later, with non-Euclidean geometries, which do not obviously model space as we experience it. Non-existence does not imply inconsistency. But any interesting abstract construct had better be internally consistent.

Second, checking the internal consistency of all of the theorems produced so far is typically not a valid proof of consistency, because (interesting) axiomatic systems generate an infinite number of theorems. The proof of consistency must guarantee that not a single theorem, including some that we have not yet produced and that we might never produce, contradicts any other theorem in the system.

One possible way to prove the consistency of an axiomatic system is model-based, where a model is a kind of interpretation.

## Monday, May 20, 2013

### Consistency of axiomatic systems

The "problem of consistency" is the topic of Section II of Nagel and Newman's book.  This section defines "consistency" and explains when and why it became an important property of axiomatic systems.

The oldest and most famous axiomatic system is that of Euclid, in which he systematized all of the knowledge of geometry (and more) available to him over two thousand years ago. Based on five axioms, Euclid was able to rigorously prove a very large number of known and new theorems (called "propositions" in his Elements). His axioms were supposed to be intuitively true. The first four axioms dealt with line segments, lines, circles and angles (see this Wikipedia entry) and have been viewed as self-evident. In contrast, the fifth axiom, which was equivalent to the following statement: "Through a point outside a given line, only one parallel to the line can be drawn" (page 9), was not intuitively true (apparently because the two lines involved extend to infinity in two directions, similarly to asymptotes).

Since this proposed axiom was not obviously true, many mathematicians tried to prove that it logically follows from the first four axioms. Only in the nineteenth century was it demonstrated that it is NOT possible to prove the parallel axiom from the first four axioms.

This proof that it is impossible to prove a given statement is a great precursor of Gödel's incompleteness theorems.

## Friday, May 17, 2013

### Formal systems, axioms, inference rules, formal proofs

I'll start this post by describing a simple formal system that I made up.

formal system is comprised of axioms and inference rules. Each axiom and inference rule is defined syntactically, that is, by ordered sequences of symbols that follow strict syntactic rules but do not (necessarily) have any meaning. The set of all symbols allowed in a formal system are explicitly listed in its alphabet, simply, a finite, non-empty set of symbols.

My formal system, let's call it FS,  uses the alphabet {E,0,1} and has only one axiom and two inference rules. Here is the axiom (in the box below):

## Wednesday, May 15, 2013

### Let's get started with Nagel and Newman (Section I)

Let's ease our way into this. I haven't read the Nagel and Newman book in a few years. But I remember it as a short and highly readable overview of the proof. I have a 1986 softcover edition by NYP Press. Today, I want to write about Section I - Introduction. It's really short: under 5 pages.

First, a short quote that makes me feel better about my past failures at REALLY understanding the proof of GIT:

## Tuesday, May 14, 2013

### I just want to understand the proof of Gödel's incompleteness theorems

So this is it! Summer 2013... This is when I get to master the proof of Gödel's incompleteness theorems (thereafter: GIT).

I discovered GIT as an undergraduate student in one of my Artificial Intelligence courses. My instructor was reading Gödel, Escher, Bach and was raving about the book. He even took some of our exam problems from it. But what I remember most is the outline of the proof of GIT, more precisely the first theorem. To be honest, the only part of it that stuck with me was the idea of numbering the formulas. Of course, at the time, I bought a copy of the book. I liked many parts of it but I never finished it.

Years later, while I was teaching computability theory and other computer science topics, I encountered GIT several times. At some point, I decided I wanted to understand the proof of GIT.

First, I read the classic "Gödel's Proof" by Nagel and Newman. That is a concise and user-friendly overview of the proof. But it was only a proof sketch. So I looked at some textbooks on formal logic, which were neither concise nor user-friendly. I still did not understand the proof.

Finally, a couple of years ago, I came across a promising volume: "Introduction to Gödel’s Theorems" by Peter Smith. I read the first few chapters but then ran out of time (and motivation) to finish it. The following year, I made another attempt: I had to start from scratch and did not get any farther into it the second time around...


## Weighted Quick Union (WQU)

Improving on Quick Union relies on a key insight: whenever we call find, we have to climb to the root of a tree. Thus, the shorter the tree the faster it takes!

New rule: whenever we call connect, we always link the root of the smaller tree to the larger tree.

Following this rule will give your trees a maximum height of $\log N$, where N is the number of elements in our Disjoint Sets. How does this affect the runtime of connect and isConnected?

Let's illustrate the benefit of this with an example. Consider connecting the two sets T1 and T2 below:

We have two options for connecting them:

The first option we link T1 to T2. In the second, we link T2 to T1.

The second option is preferable as it only has a height of 2, rather than 3. By our new rule, we would choose the second option as well because T2 is smaller than T1 (size of 3 compared to 6).

We determine smaller / larger by the number of items in a tree. Thus, when connecting two trees we need to know their size (or weight). We can store this information in the root of the tree by replacing the -1's with -(size of tree). You will implement this in Lab 6.

#### Maximum height: Log N

Following the above rule ensures that the maximum height of any tree is Θ(log N). N is the number of elements in our Disjoint Sets. By extension, the runtimes of connect and isConnected are bounded by O(log N).

Why log N? The video above presents a more visual explanation. Here's an optional mathematical explanation why the maximum height is $\log_2 N$. Imagine any element $x$ in tree $T1$. The depth of $x$ increases by $1$ only when $T1$ is placed below another tree $T2$. When that happens, the size of the resulting tree will be at least double the size of $T1$ because $size(T2) \ge size(T1)$. The tree with $x$ can double at most $\log_2 N$ times until we've reached a total of N items ($2^{\log_2 N} = N$). So we can double up to $\log_2 N$ times and each time, our tree adds a level $\rightarrow$ maximum $\log_2 N$ levels.

You may be wondering why we don't link trees based off of height instead of weight. It turns out this is more complicated to implement and gives us the same Θ(log N) height limit.

### Summary and Code

Implementation Constructor connect isConnected
QuickUnion Θ(N) O(N) O(N)
QuickFind Θ(N) Θ(N) Θ(1)
QuickUnion Θ(N) O(N) O(N)
Weighted Quick Union Θ(N) O(log N) O(log N)

N = number of elements in our DisjointSets data structure


# Euler equations for the incompressible fluids

Incompressible fluid with constant density ρ fills the three-dimensional domain below the free surface $z = η(r)$ in cylindrical polar coordinates. The flow is axisymmetric and steady, and the only non-zero velocity component is $u_θ$. Gravity acts upon the fluid. The fluid in $r\lt a$ rotates rigidly about the z-axis with angular velocity $\Omega$ and the fluid $r \ge a$ is irrotational.

Use the radial and vertical components of the Euler equations to show that the pressure $p$ in the region $z < η$, $r < a$ satisfies

$$\frac{p}{\rho} = \frac{1}{2}{Ω^2}{r^2} −gz + \text{constant}$$ and find the constant.

$$\frac{∂u_r}{∂t}+\left(u_r\frac{∂}{∂r}+u_z\frac{∂}{∂z}\right)u_r-\frac{(u_θ)^2}{r}+\frac{1}{\rho}\frac{∂p}{∂r}=0$$

$$\frac{∂u_θ}{∂t}+\left(u_r\frac{∂}{∂r}+u_z\frac{∂}{∂z}\right)u_θ+\frac{u_θu_r} {r}=0$$

$$\frac{∂u_z}{∂t}+\left(u_r\frac{∂}{∂r}+u_z\frac{∂}{∂z}\right)u_z+\frac{1}{\rho}\frac{∂p}{∂z}=-g$$

I have the Euler equation in cylindrical coordinate. Then how should I figure out the $\Omega$. The next question is to show that the free surface position in $r<a$ is $$η=\frac{\Omega^2a^2}{g}(\frac{r^2}{2a^2}-1)$$ if it is helpful. Thank you so much!

• Are you sure the radial equations aren't available to you? That's a fair amount of effort to derive those. May 15 '17 at 19:14

Sincerely, if you derived the Euler equations for cylindrical coordinates, this is a very simple exercise for you.

With the hypothesis, many terms disappear, those involving $u_r=0,\;u_z=0$ and the partials wrt time (steady flow).

Further, the motion is rigid, meaning that all fluid parts have zero relative motion. If they are moving in circles, the angular velocity for all point in the fuid $\Omega$ is the same! You don't need to figure anything, it must be known. With all this, $u_\theta=\Omega r$

$\begin{cases} -\dfrac{(u_θ)^2}{r}+\dfrac{1}{\rho}\dfrac{\partial p}{\partial r}=0\\ \dfrac{1}{\rho}\dfrac{\partial p}{\partial z}=-g \end{cases}$

Integrating the second,

$\dfrac{p}{\rho}=-gz+f(r)\tag 1$

So is $\dfrac{1}{\rho}\dfrac{\partial p}{\partial r}=f'(r)$ or $\dfrac{p}{\rho}+C=f(r)$. With the first equation,

$-\dfrac{(u_θ)^2}{r}+f'(r)=0$ or $-\dfrac{(\Omega r)^2}{r}+f'(r)=0$ or

$-\Omega^2r+f'(r)=0$. Integrating, $f(r) = \dfrac{1}{2}{Ω^2}{r^2}+h(z)$

$\dfrac{p}{\rho}= \dfrac{1}{2}{Ω^2}{r^2}+h(z)-C\tag 2$

Comparing $(1)$ and $(2)$ we have that $h(z)=-gz+C$

$\dfrac{p}{\rho} = \dfrac{1}{2}{Ω^2}{r^2} −gz +C$

I could not recover the given solution for the free surface, although I've found some relation that produces the well known paraboloid for the free surfce of the Newton's bucket. To find $C$, we can consider the heigh $z$ for the free surface ($p=0$) at some $r$. We can set $z=0$ for $r =a$.

$0=\dfrac{1}{2}{Ω^2}{a^2}+C$ or $C=-\dfrac{1}{2}{Ω^2}{a^2}$

$\dfrac{p}{\rho} = \dfrac{1}{2}{Ω^2}{r^2} −gz-\dfrac{1}{2}{Ω^2}{a^2}=\dfrac{1}{2}Ω^2(r^2-a^2)-gz$

For the free surface $z=\eta(r)$, $p=0$

$\eta=\dfrac{1}{2g}Ω^2(r^2-a^2),\;r\lt a$

From other considerations, $\eta$ for $r\ge a$ is $\eta=-\dfrac{\Omega^2a^4}{2gr^2}$. As $\eta$ has to be continuous $\eta(a)=-\dfrac{\Omega^2a^2}{2g}$, for the formula we've found for $r\lt a$

$0=\dfrac{Ω^2a^2}{2}+\dfrac{\Omega^2a^2}{2}+C$ leading to $C=\Omega^2a^2$ and to $\eta=\dfrac{\Omega^2a^2}{g}(\dfrac{r^2}{2a^2}-1)$ for $r\lt a$

$$\eta= \begin{cases} \dfrac{\Omega^2a^2}{g}(\dfrac{r^2}{2a^2}-1)& x\lt a\\ -\dfrac{\Omega^2a^4}{2gr^2}&a\le x \end{cases}$$

I've sketched the function for the free surface (setting $\dfrac{\Omega^2a^2}{2g}=1$ and $a=1$)

• It is actually the next two question of this one. math.stackexchange.com/questions/2280167/… I try to get the constant and the free surface and the number matches when I set r=a, p=0 and use the free surface when r>=a. I get C=-1/2(\Omega) ^2a^2-(\Omega)^2a^2/2 and free surface as above. Can I just plug in the limit May 16 '17 at 7:40
• And if I want to draw a sketch for the free surface displacement then it is gonna be a paraboloid but when happen at r=a then? May 16 '17 at 7:49
• @stedmoaoa, the resultig function is smooth. May 16 '17 at 13:47
• Thank you so much it really makes sense. So the graph is like x^2-1 and -1/x^2 Then, how should I describe at r=a? May 16 '17 at 14:08
• Continuous and differentiable for all quantites. May 16 '17 at 17:37


# A statement about divisibility of relatively prime integers

I'm solving a problem, and the solution makes the following statement: "The common difference of the arithmetic sequence 106, 116, 126, ..., 996 is relatively prime to 3. Therefore, given any three consecutive terms, exactly one of them is divisible by 3." Why is this statement true? Where does it come from? Is it generalizable to other numbers?

• And here I thought it was something about "relatively prime statements" – Yuriy S Jul 21 '16 at 21:56

There is a simple, well-known theorem about that:

If $a$, $b$ and $m$ are integers, $m\ge 2$ and $\gcd(b,m)=1$ then the set $\{a,a+b,a+2b,\ldots,a+(m-1)b\}$ is a complete residue system.

This implies, for example, that the set contains exactly one multiple of $m$.

• Also, most of the other answers hint at a proof of that theorem. – Jeppe Stig Nielsen Jul 22 '16 at 8:04

Take $a\in\mathbb{N}$ and suppose that $a\equiv 1\pmod 3$. Then \begin{align*} a&\equiv 1\pmod 3\\ a+10\equiv 11&\equiv 2\pmod 3\\ a+20\equiv 21&\equiv 0\pmod 3 \end{align*}

If we instead have $b\in\mathbb{N}$ such that $b\equiv 2\pmod 3$, then adding $10$ to $b$ is sufficient. The case in which we have $\equiv 0\pmod 3$ is obviously trivial.

If $\,b\,$ is coprime to $\,m\,$ then $\,a+ib,\, i = 1,\ldots,m\,$ is a complete residue system mod $\,m\,$ since

$$\ a+ib \equiv a+j b \iff (i\!-\!j)b\equiv 0\overset{(b,m)=1}\iff i\!-\!j\equiv 0\iff i = j\,\ {\rm by}\,\ 1\le i,j\le m$$

Hence, being complete, it contains an element $\equiv 0,\,$ i.e. a multiple of $\,m.$

Remark $\$ When $\,b = 1\,$ we obtain the well-known special case that any sequence of $\,m\,$ consecutive integers contains a multiple of $\,m.$

If $d$ is relatively prime to $b$, then $a + d j$ is divisible by $b$ if and only if $j \equiv -a d^{-1} \mod b$. Thus exactly one of any $b$ consecutive terms of the arithmetic sequence $a, a+d, a+2d, \ldots$ is divisible by $b$.

In general let $d$ be the common difference of an arithmetic sequence, where $3\nmid d$, and let $a$, $a+d$, and $a+2d$ be three consecutive terms in that sequence. Now WLOG if we assume $3\nmid a$ then we can write $a$ and $d$ as: $$a=3k+1\quad\text{ or }\quad a'=3k+2\quad\text{ and }\quad d=3j+1\quad\text{ or }\quad d'=3j+2$$ for some $j$, $k\in\mathbb{Z}$, where primes have been added to distinguish the two possibilities for $a$ and $d$. Now look at the four possibilities we have for $a+d$, and $a+2d$: $$a+d=3(k+j)+2,\quad \quad a+2d=3(k+2j)+3\qquad (A)$$ $$a+d'=3(k+j)+3,\quad \quad a+2d'=3(k+2j)+5\qquad (B)$$

$$a'+d=3(k+j)+3,\quad \quad a'+2d=3(k+2j)+5\qquad (C)$$ $$a'+d'=3(k+j)+4,\quad \quad a'+2d'=3(k+2j)+6\qquad (D)$$

As can be seen, since we assumed $3\nmid a$, only one member of the three consecutive terms is divisible by $3$ on listing all possible combinations (A), (B), (C), and (D). It would make no difference if we assumed $3\nmid a+d$ or $3\nmid a+2d$ as our starting point as the problem is cyclic modulo $3$. By this look at the congruences of $a$, $a+d$, $a+2d\pmod{3}$ for each of (A), (B), (C), and (D) and you see we get the full set of residue classes $[0]$, $[1]$, $[2]$ modulo $3$ in some order, which is the reason we get one and only one number divisible by $3$.

Further for some $m$ for which $\gcd(m,d)=1$ look at $m$ consecutive terms of an arithmetic sequence module $m$: $$a,\ a+d,\ a+2d,\dotsc,\ a+(m-1)d\pmod{m}\qquad (E)$$ then $a+kd\equiv a+jd\pmod{m}$ iff $(k-j)d\equiv 0\pmod{m}$ for $0\le k,j\le m-1$, which since $[0]$, $[1]$, $[2],\dotsc,\ [m-1]$ form a complete residue system modulo $m$ we must have $[k]=[j]$ modulo $m$, and so in fact $k=j$ with respect to our $m$ consecutive terms with any distinct pair of terms incongruent modulo $m$, and only one unique number in the set being divisible by $m$. To see this imagine (E) rewritten with respect to residue classes modulo $m$ reordered so: $$a+[0],\ a+[1],\ a+[2],\dotsc,\ a+[m-1]\pmod{m}\qquad (E')$$ If $m\mid a$ then $m\equiv a+[0]\pmod{m}$, and we also see $m$ cannot divide any other number in the set; if $m\nmid a$ then we can write $a=mj+k$ where $j$, $k\in\mathbb{Z}$, and $\gcd(k,m)=1$, which is in the class $[mj+k]=[k]=[k']$ modulo $m$, and $[k']+[k'']=[0]$ modulo $m$, $0<k',k''\le m-1$, where $[k']$ and $[k'']$ are additive inverses in the complete residue system (note that $[0]$ is self inverse wrt addition).


# Math Help - Counting sides

1. ## Counting sides

A rectangle that measures 4cm by 6cm is divided into 24 squares with sides 1cm in length. What is the total number of 1cm long sides in those 24 squares? (If 2 squares share a side, the side should be counted only once).

Now I know the answer is 58 and I drew a sketch and counted the sides but I was wondering if there was any shortcut to this problem that I was missing..? Setting up any kind of expression or equation or maybe a combinations problem of some sort?

2. Let use a vertical layout like this (for explanation):

****
****
****
****
****
****

First count all the right and bottom sides of each square: 24 * 2 = 48. There is no overlap in this counting.
However, we missed some sides with this count. Which ones? The entire top row and left column.
Well, the number of edges we did not count is:
# on top + # on left = 4 + 6 = 10

Grand total is: 48 + 10 = 58

So from this the general solution for number of sides in a n by m block is:
2nm + n + m = n(1 + m) + m(1 + n)

This suggest thats an alternative way is to count:
#horizontal edges + #vertical edges = n(1 + m) + m(1 + n)

3. Hi snow--your solution definitely seems to work but I think there is a column of 6 missing Thanks!

4. Hello, sarahh!

A rectangle that measures 4cm by 6cm is divided into 24 squares with sides 1cm in length.
What is the total number of 1cm long sides in those 24 squares?
(If 2 squares share a side, the side should be counted only once).

You made a sketch but you didn't see any pattern, did you?

Code:

* - * - * - * - * - * - *
|   |   |   |   |   |   |
* - * - * - * - * - * - *
|   |   |   |   |   |   |
* - * - * - * - * - * - *
|   |   |   |   |   |   |
* - * - * - * - * - * - *
|   |   |   |   |   |   |
* - * - * - * - * - * - *

The unit squares are in 4 rows and 6 columns.

Imagine these 24 squares formed with matches.
How many matches are used?

We see that there are 6 matches in each row.
And there are 4 + 1 rows.

Hence, there are: .6(4+1) horizontal matches.

We see that there are 4 matches in each column.
And there are 6 + 1 columns.

Hence, there are: .4(6 + 1) vertical matches.

We can generalize this problem.
Suppose there are R rows and C columns.

Then there are: .C(R+1) horizontal matches
. . . . . . . .and: .R(C+1) vertical matches.

Total: . $C(R+1) + R(C+1) \;=\; R + C + 2RC$ matches.

5. Originally Posted by sarahh
Hi snow--your solution definitely seems to work but I think there is a column of 6 missing Thanks!
There is a row of 6 and a column of 4 missing. Which is why a 10 (= 6 + 4) is added at the end.

6. Just for fun, here is another approach-- first do it the wrong way, then patch it up.

There are 24 squares in all, each with 4 sides, and each side is shared between two squares, so there are

$\frac{4 \times 24}{2} = 48$ sides.

But oops, the sides along the perimeter of the rectangle (2 * (4 + 6) = 20) are not shared! So we need to add those back in.

$48 + \frac{20}{2} = 58$


Fibonacci's final digits cycle every 60 numbers

How would you go about to prove that the final digits of the Fibonacci numbers recur after a cycle of 60?

References:

The sequence of final digits in Fibonacci numbers repeats in cycles of 60. The last two digits repeat in 300, the last three in 1500, the last four in , etc. The number of Fibonacci numbers between and is either 1 or 2 (Wells 1986, p. 65).

http://mathworld.wolfram.com/FibonacciNumber.html

• Why do you conjecture that? Do you have a reference that asserts the statement? – Giovanni De Gaetano Feb 26 '12 at 9:52
• I added some references. – crazyGuy Feb 26 '12 at 9:57
• Neat! Have a look at this link. – Juan S Feb 26 '12 at 10:00
• Here is a link: oeis.org/A096363 – Stefan Gruenwald Dec 14 '18 at 20:59

Notice that:

$$F_{n+15} \equiv 7F_n \pmod{10}$$ for $$n\geq 1$$.

Also the order of $$7$$ mod $$10$$ is $$4$$ so the repetition in the digits of the Fibonacci numbers begins after place $$15\times 4 = 60$$.

• I got that recursion by just writing out the first $16$ Fibonacci numbers mod $10$ and noticing that you get $1,1,...,7,7,...$, this dictates that that next $13$ numbers must be $7$ times their corresponding numbers (mod $10$) in the first lot of $15$ numbers. – fretty Feb 26 '12 at 10:03
• How do we know that F_n+15 = 7F_n mod 10 ? and How do you get that 7 mod 10 is 4 ? thanks. – crazyGuy Feb 26 '12 at 10:07
• Write out the first 16 Fibonacci numbers mod 10: 1,1,...,7,7,... Now by the recursion of the fibonacci numbers the fact that this "multiple-repeat" has happened now tells us that the next 13 numbers must be the same as the 3rd-15th Fibonacci numbers but multiplied by $7$ mod 10. So an actual repeat will happen as soon as we have multipled by enough $7$'s to get $1$ mod 10. This is the "order" of $7$ mod $10$. It is easy to work out here, $7^2 \equiv 4$ mod $10$ and $7^4 \equiv 1$ mod $10$ so the order is $4$. – fretty Feb 26 '12 at 10:13
• @sic2 He's not saying that $7\mod10$ is $4$, but rather that $7^4\equiv 1\mod 10$ and that this is not true for any smaller power of $7$, which is a consequence of Euler's theorem. – Alex Becker Feb 26 '12 at 10:15
• Also this method generalises to the Fibonacci numbers mod $n$. Find the first $1,1,...,a,a,...$ then work out the order of $a$ mod $n$ and you will have your first place where it recurs. – fretty Feb 26 '12 at 10:29

Since each term in the Fibonacci sequence is dependent on the previous two, each time a $0\pmod{m}$ appears in the sequence, what follows must be a multiple of the sequence starting at $F_0,F_1,\dots=0,1,\dots$ That is, a subsequence starting with $0,a,\dots$ is $a$ times the sequence starting with $0,1,\dots$

Consider the Fibonacci sequence $\text{mod }2$: $$\color{red}{0,1,1,}\color{green}{0,1,1,\dots}$$ Thus, the Fibonacci sequence repeats $\text{mod }2$ with a period of $3$.

Consider the Fibonacci sequence $\text{mod }5$: $$\color{red}{0,1,1,2,3,}\color{green}{0,3,3,\dots}$$ Thus, the Fibonacci sequence is multiplied by $3\pmod{5}$ each "period" of $5$. Since $3^4=1\pmod{5}$, the Fibonacci sequence repeats $\text{mod }5$ with a period of $20=4\cdot5$.

Thus, the Fibonacci sequence repeats $\text{mod }10$ with a period of $60=\operatorname{LCM}(3,20)$.

Note that

$$\begin{pmatrix} F_{n+1}\\ F_{n+2} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} F_n \\ F_{n+1} \end{pmatrix}$$

and

$$\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^{60} \equiv \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \mod 10.$$

One can verify that $60$ is the smallest power for which this holds, so it is the order of the matrix mod 10. In particular

$$\begin{pmatrix} F_{n+60}\\ F_{n+61} \end{pmatrix} \equiv \begin{pmatrix} F_n \\ F_{n+1} \end{pmatrix} \mod 10$$

so the final digits repeat from that point onwards.

• The same matrix applies to the Lucas Numbers, however, $\text{ mod }10$, they have a period of $12$. The fact that $\begin{pmatrix} 0&1\\1&1\end{pmatrix}^{60}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\text{ mod }10$, says that the period of a sequence satisfying $a_n=a_{n-1}+a_{n-2}$ divides $60$. – robjohn Feb 26 '12 at 19:21

$F_{0}=F_1=F_{60}=F_{61}= 1 \mod10$

By inspection, these are the first two pairs of consecutive Fibonaccis for which this is true. Since the recurrence relation only takes into account the previous two terms and last digits only depend on previous last digits, this suffices to prove the claim.

I have found that there is an 11 number grouping. You must reduce numbers to a single digit (13=4 or 55=10=1). After 24 numbers into the sequence it repeats. To go a little deeper, a 9 is in the twelfth and twenty- fourth sequence and the real number is also divisible by 12 in these positions (which coincidentally is every twelfth number in the sequence).

Now if you do the multiplication table of numbers 1 through 8 and reduce all numbers to a single digit, you will find that 1 and 8 correspond in reverse with each other. The same for 2 and 7, also 4 and 5. 3 and 6 are different. You will notice that for every number sequence it will repeat after a 9 appears. So, Fibonacci sequence would be this:

1,1,2,3,5,8,4,3,7,1,8,9,8,8,7,6,4,1,5,6,2,8,1,9,1,1,2,3,5,8,4,3,7,1,8,9,8,8,7,6,4,1,5,6,2,8,1,9......

Remember how I said 1 and 8, 2 and 7, 4 and 5 correspond with each other by reducing the multiplication table to single digits and all numbers repeat a sequence after a 9?

If you move the eleven number sequence between the nines and set 1,1,2,3... above the sequence 8,8,7,6... You will find that they all correspond with the reducing method afore mentioned.

It's very interesting.

• What do you mean $11$-number grouping? Why are you reducing mod $9$ instead of mod $10$? How does this prove that it repeats with cycle $60$? – 6005 Oct 24 '16 at 2:58
• I apologize. This was not to prove or disprove a 60 cycle repeat. Maybe I posted in the wrong area? I just found this sequence repeat and I thought I would give it to bigger brains and see what they can make of it. – Michael Bunton Oct 24 '16 at 3:12
• 11 number grouping means the first 11 numbers in the sequence. The twelfth number is 144 and that reduces to 9 i.e.(1+4+4=9). Then a second set of eleven numbers followed by a reduced 9. – Michael Bunton Oct 24 '16 at 3:20


# find Limit $a_n= \frac{1}{n+1}+\frac{1}{n+2}+…+\frac{1}{n+n}$ [duplicate]

show sequence $a_n= \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}$ converges to log2

my attempt:

1. sequence $a_n$ is monotonic increasing
2. 0<$a_n$<1/2

how to find limit?

• Oh come on, this must be the duplicate of three duplicates. – Vincenzo Oliva Aug 14 '15 at 9:39

$$\lim_{n\rightarrow \infty}\sum_{i=1}^{n}\frac 1 {n+i}=\lim_{n\rightarrow \infty}\sum_{i=1}^{n}\frac 1 n\frac 1 {1+\dfrac in}=\int_0^1\frac 1 {1+x}dx=\ln(1+1)-\ln(1+0)=\ln(2)$$ (Riemann sum)

• +1, but there should be a limit as $n \to \infty$ instead of the sum going to infinity. The sum goes fom $i=1$ to $i=n$ – 6005 Aug 14 '15 at 9:07
• @6005 Fixed, thank you! – GeorgSaliba Aug 14 '15 at 9:09
• ya i was hoping there would be some other way. riemann sum cannot be used to compute limit of sequence$a_n=1+1/2+1/3...+1/n−log(n+1)$ – ketan Aug 14 '15 at 9:19
• @ketan Then why not use the harmonic number $H_n$ that was already mentioned by @MichaelGaluza? But I suppose this becomes a slightly different question than the one you asked. – GeorgSaliba Aug 14 '15 at 9:21
• @ketan $a_n=1+1/2+1/3+...+1/n - \log(n+1)=H_n -\log(n+1)=\log(n)+\gamma + \epsilon_n-\log(n+1)=\log(n/(n+1))+\gamma +\epsilon _n$ – GeorgSaliba Aug 14 '15 at 9:23

\begin{align} \log(2) &=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\\ &=\lim_{n\to\infty}\sum_{k=1}^{2n}\frac{(-1)^{n-1}}k\\ &=\lim_{n\to\infty}\left(\vphantom{\sum_{k=1}^n}\right.\overbrace{\sum_{k=1}^{2n}\frac1k}^{\substack{\text{sum of all}\\\text{terms}}}-2\overbrace{\sum_{k=1}^n\frac1{2k}}^{\substack{\text{sum of even}\\\text{terms}}}\left.\vphantom{\sum_{k=1}^n}\right)\\[6pt] &=\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1k\right)\\[6pt] &=\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac1k\\ \end{align}

• Nice overbrace. Is there a corresponding underbrace command or somethings similar? – mathreadler Aug 14 '15 at 9:23
• Yep. It is \underbrace :-) Then you use subscripts instead of superscripts. – robjohn Aug 14 '15 at 9:23

Here is classical proof of it. Let $H_n=\sum_{i=1}^n 1/i$. It's well-known that $$H_n = \ln n + \gamma + \epsilon_n,$$ where $\gamma$ is Euler–Mascheroni constant, $\epsilon_n\to 0$. So, $$a_n = H_{2n}-H_{n} = \ln 2 + \epsilon_{2n} - \epsilon_n \to \ln2.$$

• what is $\epsilon_n$? is it always zero? – ketan Aug 14 '15 at 9:28
• @ketan, no. It's sequence which tends to $0$. – Michael Galuza Aug 14 '15 at 11:39

Notice, $$a_n=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\ldots +\frac{1}{n+n}$$ $$=\sum_{r=1}^{\infty}\frac{1}{n+r}=\sum_{r=1}^{\infty}\frac{1}{n\left(1+\frac{r}{n}\right)}$$$$=\sum_{i=1}^{\infty}\frac{\frac{1}{n}}{1+\frac{r}{n}}$$ Now, let $\frac{r}{n}=x\implies \frac{1}{n}=dx\to 0$ $$\text{upper limit of x}=\lim_{n\to \infty}\frac{n}{n}=1$$ $$\text{lower limit of x}=\lim_{n\to \infty}\frac{1}{n}=0$$ Hence, using integration we get $$a_n=\int_{0}^{1}\frac{dx}{1+x}$$ $$=\left[\ln (1+x)\right]_{0}^{1}$$ $$=\left[\ln 2-\ln 1\right]=\ln 2$$


# Math Help - Proving the binomial coeficient is always a natural number?

1. ## Proving the binomial coeficient is always a natural number?

How would one go about proving that:

${ n \choose k }$

is always a natural number? That is, assuming $0 \leq k \leq n$ and that $n = 1, 2, 3, 4...$ and that $k = 1, 2, 3, 4...$; what I mean by that is that $n$ and $k$ are both positive integers. Its a question in my textbook that I bought early to get ahead on fall classes, so I don't have any teachers to ask about this question; not till fall atleast. I'm not asking for a straight out proof, but preferably some direction as to where I would go, which direction I should go to prove this? A litttle guidence is all I need, please don't simply give me the full answer, if you don't mind. Thanks in advance

2. Originally Posted by mfetch22
How would one go about proving that:

${ n \choose k }$

is always a natural number? That is, assuming $0 \leq k \leq n$ and that $n = 1, 2, 3, 4...$ and that $k = 1, 2, 3, 4...$; what I mean by that is that $n$ and $k$ are both positive integers. Its a question in my textbook that I bought early to get ahead on fall classes, so I don't have any teachers to ask about this question; not till fall atleast. I'm not asking for a straight out proof, but preferably some direction as to where I would go, which direction I should go to prove this? A litttle guidence is all I need, please don't simply give me the full answer, if you don't mind. Thanks in advance
Use the fact that $\displaystyle{\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$ for $\displaystyle{k=1\ldots n-1}$ and $\displaystyle{\binom{n}{0}=\binom{n}{n}=1}$.

(This will be a proof by induction)

3. Originally Posted by mfetch22
How would one go about proving that:

${ n \choose k }$

is always a natural number? That is, assuming $0 \leq k \leq n$ and that $n = 1, 2, 3, 4...$ and that $k = 1, 2, 3, 4...$; what I mean by that is that $n$ and $k$ are both positive integers. Its a question in my textbook that I bought early to get ahead on fall classes, so I don't have any teachers to ask about this question; not till fall atleast. I'm not asking for a straight out proof, but preferably some direction as to where I would go, which direction I should go to prove this? A litttle guidence is all I need, please don't simply give me the full answer, if you don't mind. Thanks in advance
There's also the combinatorial proof. (The binomial coefficient is given as an "archetypal example" not far from the top of the page.) The formula(s) given for $\displaystyle { n \choose k }$ can be shown to equal either of the two equivalent quantities

1) the coefficient of the $\displaystyle x^k$ term in the polynomial expansion of the binomial power $\displaystyle (1 + x)^n$
2) the number of ways to choose k elements from a set of n elements. (The number of distinct k-subsets of {1, 2, ... , n}.)

Obviously both of these quantities are integers.

4. Another way is to use Legendre's formula, which states that ( $\mathbb{P}$ is the set of prime numbers)

$
n! = \prod_{p\in \mathbb{P}} p^{\sum_{i=1}^\infty \Big\lfloor\frac{n}{p^i}\Big\rfloor}
$

Thus for any prime $p$ we have:

$
\binom{n}{k} = \frac{n!}{k!(n-k)!} = \prod_{p\in \mathbb{P}} p^{\sum_{i=1}^\infty \Big\lfloor\frac{n}{p^i}\Big\rfloor-\Big\lfloor\frac{k}{p^i}\Big\rfloor-\Big\lfloor\frac{n-k}{p^i}\Big\rfloor}}
$

So all what's left is to show that $\lfloor{x+y}\rfloor \ge \lfloor{x}\rfloor + \lfloor{y}\rfloor$

so it follows that for all $i\in\mathbb{N}$:

$
\Big\lfloor\frac{n}{p^i}\Big\rfloor-\Big\lfloor\frac{k}{p^i}\Big\rfloor-\Big\lfloor\frac{n-k}{p^i}\Big\rfloor} \ge 0
$

and therefore $\binom{n}{k} \in \mathbb{N}$

5. Originally Posted by Unbeatable0
Another way is to use Leibnitz' formula, which states that ( $\mathbb{P}$ is the set of prime numbers)

$
n! = \prod_{p\in \mathbb{P}} p^{\sum_{i=1}^\infty \Big\lfloor\frac{n}{p^i}\Big\rfloor}
$

Thus for any prime $p$ we have:

$
\binom{n}{k} = \frac{n!}{k!(n-k)!} = \prod_{p\in \mathbb{P}} p^{\sum_{i=1}^\infty \Big\lfloor\frac{n}{p^i}\Big\rfloor-\Big\lfloor\frac{k}{p^i}\Big\rfloor-\Big\lfloor\frac{n-k}{p^i}\Big\rfloor}}
$

So all what's left is to show that $\lfloor{x+y}\rfloor \ge \lfloor{x}\rfloor + \lfloor{y}\rfloor$

so it follows that for all $i\in\mathbb{N}$:

$
\Big\lfloor\frac{n}{p^i}\Big\rfloor-\Big\lfloor\frac{k}{p^i}\Big\rfloor-\Big\lfloor\frac{n-k}{p^i}\Big\rfloor} \ge 0
$

and therefore $\binom{n}{k} \in \mathbb{N}$
Perfect , in fact $\lfloor{ x+y \rfloor} = \lfloor{ x \rfloor} + \lfloor{ y \rfloor} + \lfloor{ \{ x \} + \{ y \} \rfloor} \geq \lfloor{ x \rfloor} + \lfloor{ y \rfloor}$ .

6. Originally Posted by mfetch22
How would one go about proving that:

${ n \choose k }$

is always a natural number? That is, assuming $0 \leq k \leq n$ and that $n = 1, 2, 3, 4...$ and that $k = 1, 2, 3, 4...$; what I mean by that is that $n$ and $k$ are both positive integers. Its a question in my textbook that I bought early to get ahead on fall classes, so I don't have any teachers to ask about this question; not till fall atleast. I'm not asking for a straight out proof, but preferably some direction as to where I would go, which direction I should go to prove this? A litttle guidence is all I need, please don't simply give me the full answer, if you don't mind. Thanks in advance
If you are asking from the perspective of the Binomial Expansion,
then you can observe a pattern if you refrain from expressing the terms with indices...

$(a+b)^2=(a+b)(a+b)=aa+ab+ba+bb$

"a" times "b" is expressed in both possible ways.

$(a+b)^3=(a+b)(a+b)^2=(a+b)(aa+ab+ba+bb)$

$=aaa+aab+aba+abb+baa+bab+bba+bbb$

$=aaa+aab+aba+baa+abb+bab+bba+bbb$

If the "a" values were labelled $a_1,\ a_2,\ a_3$ we could arrange them,

hence we have all possible selections of 2 "a"'s with a "b", 2 "b"'s with an "a",
the only possible selection of 3 "a"s, the only selection of 3 "b"'s.

For higher powers, the same pattern exists.
Since this pattern is the same as "choose k from n available",
the binomial coefficient may be calculated using

$\binom{n}{k}$

There are k! arrangements of a selection.
As shown above, the number of selections is a natural number,
which can be calculated by "unarranging" the arrangements.

The number of arrangements of k from n is n(n-1)(n-2)...(n-[k-1])

which is $\frac{n!}{k!}$ as k! has been "cut off" the tail.

The number of arrangements is k! times the number of selections.