Conversation started Jan 17, 2020 at 18:33.
MyWrathAcademia
MyWrathAcademia
Jan 17, 2020 18:33
Hi @Thorgott, glad your'e here. About that count of numbers divisible by $n$ between $a$ to $b$ problem, I want to clear up some misunderstandings:
How is $k > = a/n$ the smallest $k$ when for $a = 7$ $b = 14$ $n = 6$, $k >= a/n$ would mean that the smallest $k$ is $1$ which is not true. Isn't it more correct to say that $k >= a/n$ is the lower bound for $k$ and that the actual smallest $k$ can only be found by plugging a value for $k$ starting from (i.e. at or near) this lower bound into $nk >= a$? Like wise for $k <= b/n$.
Can you also confirm that $\lceil\frac bn\rceil + 1$ doesn't literally mean $b/n - \frac an + 1$ since I think $\lceil\frac bn\rceil$ and $\lceil \frac an \rceil$ are inequalities. I was trying to write a program that can count the number of numbers from $a$ to $b$ divisible by $n$ without using any for loop.
Am I right in thinking that the formula $\lceil \frac bn \rceil - \lceil \frac an\rceil + 1$ would still require me to loop over values of $k$ in the range $\frac bn$ to $\frac an$ and compare each value to the condition $nk >= a$ and the condition $nk <= b$ in order to find the smallest $nk >= a$ and the largest $nk <= b$ in the range $b/n$ to $\frac an$?
Formatting mistake, I meant to say "Can you also confirm that $\lceil\frac bn\rceil - \lceil \frac an \rceil + 1$ doesn't literally mean $\frac bn - \frac an + 1$"
Thorgott
Thorgott
In your example, $k=1$ yields $6$, which does not lie in the range from $7$ to $14$. Yes, $\lfloor\frac{b}{n}\rfloor-\lceil\frac{a}{n}\rceil+1$ is not the same as $\frac{b}{n}-\frac{a}{n}+1$, the former expression is always integer, the latter usually not (try computing some examples). Using a loop to count expressions in unnecessary, because we've already counted them. Computing $\lfloor\frac{b}{n}\rfloor$ is the same as rounding $\frac{b}{n}$ down, which I'm sure any computer can do for you.
topologicalmagician
topologicalmagician
@Thorgott if they are empty then the index set is the empty set, right?
Thorgott
Thorgott
No?
topologicalmagician
topologicalmagician
Jan 17, 2020 18:52
@Thorgott if they are equal then $I$ is finite. If they're empty then what happens?
Thorgott
Thorgott
Here's what I mean: Let $I$ be an uncountable set and $A$ a countable set. Define $A_i=A$ for all $i\in I$. Then $(A_i)_{i\in I}$ is a collection of countable sets. Furthermore, $\bigcup_{i\in I}A_i=A$ is countable, but $I$ is uncountable.
MyWrathAcademia
MyWrathAcademia
@Thorgott I understand that $k = 1$ yields $6$. What I don't understand is for $k >= \frac an$ in the above example when $a = 7$ and $n = 6$ then the inequality $k >= 7/6$ (i.e. $k >= 1)$ means that $k$ is greater than or equal to $1$, but the minimum $k$ is $2$ for the example $a = 7$ $b = 14$ $n = 6$. Basically I'm asking why does the condition $k >= \frac an$ hold when $k = 1$ even though $k = 1$ is not the smallest $k$
Thorgott
Thorgott
It doesn't. $1\not\ge7/6$
MyWrathAcademia
MyWrathAcademia
Oh crap, sorry @Thorgott I'm used to rounding down in programming that I forgot you don't round down in mathematics
Thorgott
Thorgott
Finding the smallest natural number $k\ge\frac{a}{n}$ is the same as rounding $\frac{a}{n}$ up, so that's what you want to do.
MyWrathAcademia
MyWrathAcademia
Jan 17, 2020 19:00
Yes $ 1 \not\ge1.166666667$
So your'e saying that for example you round $1.166666667$ to $2$?
Thorgott
Thorgott
Yes. And $k=2$ is the right answer.
MyWrathAcademia
MyWrathAcademia
If you round up for $k \ge \frac an$ does that mean that you round down for $k \le \frac bn$?
I'm guessing thats why you round down for $k \ge \frac an$?
Thorgott
Thorgott
Yes, you round $\frac{a}{n}$ up and $\frac{b}{n}$ down. That's exactly what $\lceil\frac{a}{n}\rceil$ and $\lfloor\frac{b}{n}\rfloor$ mean, respectively.
MyWrathAcademia
MyWrathAcademia
Wow that makes so much sense. What are those type of brackets called?
Thorgott
Thorgott
Floor and ceiling functions
In mathematics and computer science, the floor function is the function that takes as input a real number x {\displaystyle x} and gives as output the greatest integer less than or equal to x {\displaystyle x} , denoted floor ⁡ ( x ) = ⌊ x ⌋ {\displaystyle \operatorname {floor} (x)=\lfloor x\rfloor } . Similarly, the ceiling function maps x {\displaystyle x} to the...
MyWrathAcademia
MyWrathAcademia
Jan 17, 2020 19:12
@Thank you, the celing function $\lceil \frac an \rceil$ and floor function $\lfloor \frac bn\rfloor$ are quite intuitive
@Thorgott now I understand $k \ge \frac an$ and $k \le \frac bn$ perfectly and that the former's celing function rounds up and the latter's floor function rounds down that answers my second question about what $\lceil \frac an\rceil - \lfloor \frac bn\rfloor + 1$ means
topologicalmagician
topologicalmagician
Jan 17, 2020 19:37
If $(X_i)_{j \in J}$ is a collection of $n-manifolds$ and their disjoint union is an n-manifold then $J$ is countable.
I\m still quite stuck on this
Thorgott
Thorgott
Demonstrating failure of second-countability in case of $J$ uncountable sounds like a good approach
topologicalmagician
topologicalmagician
Jan 17, 2020 19:55
@Thorgott. I can define a map $f$ $:$ $I \rightarrow X_i^*$ (where $X_i^*$ is the image of the canonical injection), the map is injective so $I$ is countable?
MyWrathAcademia
MyWrathAcademia
@Thorgott thank you. Your'e right that using a loop to count expressions is unnecessary, because we've already counted them. I have computed the expression $\lfloor \frac bn \rfloor - \lceil \frac bn \rceil + 1$ that allows me to observe that this expression has already counted the number of natural numbers divisible by $n$ between $a$ and $b$ and also that this expression is always integer.
@Thorgott your responses have been immensely helpful, thanks a lot! Is there a way to bookmark a conversation or series of comments in this chat room so that I could quickly refer to it?
 
Conversation ended Jan 17, 2020 at 19:58.