How do I know what multiplum I need?

[Nemanjavuk69]

Dear everyone


I am looking for a theorem, proof, Definition or anything else, which tells me how to find the multiplum

Suppose I have this equation $a=99$

I need to make $a$ be into the interval [150;200]. Suppose we now get told a number we can add to $a$ as many times we want as long as $a$ is inside the interval [150;200]. This number we get is 10.

We can now construct the following equation $a=99+(10 \cdot n)$ where $n$ is the multiplum of 10.

We are restricted to only allowing $n$ to be of integer type.

My question is, how do we know what $n$ has to be? It is obivious that $n$ can be [6;10]. Is there a way to calculate that $n$ can be in this interval?


Thanks in advance and have a fantastic night/ day forward

 

[Otis]

Hi. In your example using [150;200] and 10, it looks like you're seeking Integer solutions for this compound inequality:

150 ≤ 99 + 10n ≤ 200

Line Edited: To obtain Integer endpoints for your solution set, you can round the Real endpoints up and down.

Is this for schoolwork, or are you trying to write code … something else? Let us know, if I've misunderstood your question.
$\;$

 

[Nemanjavuk69]

Hi. In your example using [150;200] and 10, it looks like you're seeking Integer solutions for this compound inequality:

150 ≤ 99 + 10n ≤ 200

You can round the two solutions (i.e., the endpoints of the solution interval) to the nearest Integer.

Is this for schoolwork, or are you trying to write code … something else? Let us know, if I've misunderstood your question.
$\;$

Dear Otis

First of, thank you for replying. Second of, this is not schoolwork nor coding. This is just me pondering how, let's say, a computer would find the minimum and maximum that $n$ can be. I found this fascinating and wanted to experiment and see if there was a theorem, proof, definition, formula etc. for such.

Regarding your compound inequality, can I ask what you mean by

You can round the two solutions (i.e., the endpoints of the solution interval) to the nearest Integer.

Furthermore, here is my attempt to solve for $n$ using your approach


Would it be correct to assume, that the lower of the two end points ALWAYS uses the ceiling and the higher of the two end points ALWAYS uses the floor on the decimal number in order to find the interval for $n$?

 

[Otis]

Regarding your compound inequality, is there a way to isolate n?

Hi. Yes, using algebra, similar to how we would solve equations (like 99+10n=150 and 99+10n=200).

The inequality relates three expressions:

150 ≤ 99 + 10n ≤ 200

We carry out the same steps on each expression. First, we isolate the 10n term, by subtracting 99 and simplifying:

150 – 99 ≤ 99 + 10n – 99 ≤ 200 – 99

51 ≤ 10n ≤ 101

Now solve for n, by dividing each expression by 10 (Note: We do not change the direction of the inequality symbols because we are dividing by a positive number):

51/10 ≤ (10/10)n ≤ 101/10

Simplifying yields a solution set for all Real numbers n:

5.1 ≤ n ≤ 10.1

From that, we can see the Integer-solution's interval endpoints. Any Integer in [6;10] works.

99 + 10(6) = 159

99 + 10(10) = 199


$\;$

 

[Nemanjavuk69]

Hi. Yes, using algebra, similar to how we would solve equations (like 99+10n=150 and 99+10n=200).

The inequality relates three expressions:

150 ≤ 99 + 10n ≤ 200

We carry out the same steps on each expression. First, we isolate the 10n term, by subtracting 99 and simplifying:

150 – 99 ≤ 99 + 10n – 99 ≤ 200 – 99

51 ≤ 10n ≤ 101

Now solve for n, by dividing each expression by 10 (Note: We do not change the direction of the inequality symbols because we are dividing by a positive number):

51/10 ≤ (10/10)n ≤ 101/10

Simplifying yields a solution set for all Real numbers n:

5.1 ≤ n ≤ 10.1

From that, we can see the Integer-solution's interval endpoints. Any Integer in [6;10] works.

99 + 10(6) = 159

99 + 10(10) = 199


$\;$

Hello again Otis. I am so sorry, I think you were too fast for my post to get updated, I apologize for that. Apparently the edit did not go through so I had to update my question to you again. Do you mind checking what I wrote again?

 

[Otis]

Would it be correct to assume, that the lower of the two end points ALWAYS uses the ceiling and the higher of the two end points ALWAYS uses the floor

Yes, very good.

If a Real-solution interval endpoint is not an Integer, then it lies outside the Integer-solution set. Sketch a portion of the Real number line, as a visual aid.

If the left-endpoint is not an Integer, then it's too small to lie within the Integer-solution set (its position is too far to the left), so we round up.

If the right-endpoint is not an Integer, then it's too big (located too far to the right), so we round down.

That's what your use of the ceiling and floor functions do here; they "trim off" the parts of the Real number line that lie outside the Integer-solution's endpoints.
$\;$

 

[Nemanjavuk69]

Yes, very good.

If a Real-solution interval endpoint is not an Integer, then it lies outside the Integer-solution set. Sketch a portion of the Real number line, as a visual aid.

If the left-endpoint is not an Integer, then it's too small to lie within the Integer-solution set (its position is too far to the left), so we round up.

If the right-endpoint is not an Integer, then it's too big (located too far to the right), so we round down.

That's what your use of the ceiling and floor functions do here; they "trim off" the parts of the Real number line that lie outside the Integer-solution's endpoints.
$\;$

That makes perfect sense. Thank you so much Otis for taking your time to teach me this. May you have a wonderful day/ night forward!

 

[Otis]

you were too fast for my post to get updated

That's okay. Re-reading your edited post#3 allowed me to find a goof in my first reply. I was incorrect to say "nearest" Integer because I'd been thinking from inside the solution set you desire. (I'll get that misstatement fixed.)
$\;$

 

[pka]

I am looking for a theorem, proof, Definition or anything else, which tells me how to find the multiplum
Suppose I have this equation $a=99$
I need to make $a$ be into the interval [150;200]. Suppose we now get told a number we can add to $a$ as many times we want as long as $a$ is inside the interval [150;200]. This number we get is 10.

To: Nemanjavuk69, There are two very useful functions here, if $\bf a$ is a real number
the ceiling function of $\left\lceil\bf a \right\rceil$ is the least integer not less than $\bf a$.
Ex: $\left\lceil{79.8} \right\rceil=80~\&~\left\lceil{-9.8} \right\rceil=-9$
the ceiling function of $\left\lceil\bf a \right\rceil$ is the least integer not less than $\bf a$
And the floor function $\left\lfloor \bf a \right\rfloor$ is the largest integer not greater than $\bf a$
$\left\lfloor{79.8} \right\rfloor=79~\&~\left\lfloor{-9.8} \right\rfloor=-10$

Here two links solving this question using these functions. LINK1 & LINK2