# Thread: Need help with science/math schoolwork? Post here!

1. ## Re: Need help with science/math schoolwork? Post here!

Originally Posted by John Doe
5/4 has a left over of one, but any integer divide by four will give a leftover of 0,1,2 and 3. I'm french, so we might call leftover what the anglosaxon call the carry.

if n an m are rational, then themself could be turn into a fraction of integers.
No this is incorrect in my math vocabulary (5/4)

there exists equation and inequality which must be followed...
m/n = q*n +r (presumably also n ≠0)

0≤ r <n

q, m, n, r, must belong to integers

You could calculate it further, it is known, that 5/4, does divide "evenly" such that you get an ending decimal form. The ending result is 1,25

5/4= 1,25 x 4 + r

r=0

HOWEVER 1,25 is NOT integer. Integers are whole numbers, whether negative or positive.

Does that division equation apply only with integers, or with natural numbers? In any case, it is clear that 1,25 is not a natural number, and it is not an integer.

So, therefore it seems only logical that the remainder is the first possible natural number, which is left over, from the division of two natural numbers (in such cases where remainder is not zero.)

Therefore, the remainder of 5/4, is 1. Because 1 is the first natural number, which is left over from 5 divided by 4.

2. ## Re: Need help with science/math schoolwork? Post here!

Originally Posted by John Doe
When you divide integers n by m, n=q*m+r where r is the leftover: you know its value is 0 or any integer strictly below m, basically r belong to {0,1,2,...,m-1} a finite set of numbers. If the leftover isn't 0, then you multiply the leftover by 10 and try to divide it again by m (to get the next decimal of you result), and you repeat the process until either the leftover is 0 or of the same value of a previous leftover (start of the cycle). So at the worst case scenario, if you havent got a leftover equal to zero or equal to a previous leftover at the (m-1)th division, you will at the mth division because you have exhausted all other possible result.

I did not demonstrate that m/n is either zero repetend or repeating decimal, I demonstrated it can't be otherwise (it's obviously not very rigourous but it will do).
Assume that you would have to calculate some kind of ridiculous fractions, How can it be certain, that a cycle even exists, at all? Something like a rational number, integer divided by another integer, in long division, with pen and paper.

254178 / 7855

When n/m, and n,m belong to integers, and m is unequal to zero.

From where does it follow that the number of "leftovers" to be investigated by long division, = m-1

How can you be sure that the decimal is, actually, repeating exactly in the correct order also? This is the definition of repeating decimal, repeating decimal that repeats exactly, that some kind of cycle can be seen and defined. Does this existance of a cycle, follow, from any procedure, or formula (or axiom)?

Don't rely on "circular logic" such that irrational numbers aren't included, therefore, there must be a cycle. (if that's the real proof then silly me!)

3. When you decompose the long division of 5 by 4, the first step is that 5= 1*4+1
then you multiply the leftover, here 1, by 10 to get the next decimal number and you start again: 1*10=10=2*4+2
here the leftover (here 2), it isn't 0 or the same as before (it's not 1 again) so you carry on the procedure by multiplying the leftover by 10 and dividing the result by 4: 2*10=20=5*4+0

This time the leftover is zero therefore you stop and you actually find 1.25, which is also 125/100.... Even though 1.25 isn't an integer, what matter is that at each step we took the leftover of each division was one, and it's at the end that we wrote the result into its decimal form (aka 1.25)

Now for m/n, we have been ask to prove that the result is "either zero repetend or repeating decimal", nothing more. We don't care to know how many decimal there is if it's repetend or how many decimal there are if repeating. For this we need to focus on the property of the leftover, and as you said above:
m/n = q*n +r (presumably also n ≠0)

0≤ r <n

q, m, n, r, must belong to integers
lets do the step one m=q1*n+r1
so r1 can only be equal to only one of those integers: 0; 1; 2; 3; .....; n-2; n-1, note that there is only (n-1) possible value that isn't 0 (earlier, in 5/4 leftover could only have the value 0; 1; 2; 3)
Now if r1=0 then we're done, if not we have to divide r1 by n for step2
r1=q2*n+r2
Now if r2=0 we are done, if r2=r1 then we have entered a cycle. Else we have to divide r2 by n again for step3
r2=q3*n+r3
if r3=0 we are done, if r3 equals r1 or r2 we have entered a cycle and are done, else I'll divide r3 by n for step4

and so on and so forth

What matter here is that since there is a limited number of possible value for the leftover, I'm bound either to finally get 0 or any leftover I've already found previously. 0≤ r <n with r and n being integer are the key to prove that m/n is either zero repetend or repeating decimal (technically we proved it can't be another way)
.

for 254178 / 7855

any leftover of any step could only have the value 0; 1; 2; 3;.....; 78853; 78854.
so if after 78854 steps I haven't found a leftover that is 0 or one that I have already found, then I will at step 78855 because there isn't any other option left.

How can you be sure that the decimal is, actually, repeating exactly in the correct order also
because if the numbers are the same, so will the result of their division, what matter here is the value of ri, the value of qi is irrelevent.

From where does it follow that the number of "leftovers" to be investigated by long division, = m-1
because 0=<r<n, r and n being integers, it's easy to count how many number there are between 1 and n-1

4. ## Re: Need help with science/math schoolwork? Post here!

Originally Posted by John Doe
When you decompose the long division of 5 by 4, the first step is that 5= 1*4+1
then you multiply the leftover, here 1, by 10 to get the next decimal number and you start again: 1*10=10=2*4+2
here the leftover (here 2), it isn't 0 or the same as before (it's not 1 again) so you carry on the procedure by multiplying the leftover by 10 and dividing the result by 4: 2*10=20=5*4+0

This time the leftover is zero therefore you stop and you actually find 1.25, which is also 125/100.... Even though 1.25 isn't an integer, what matter is that at each step we took the leftover of each division was one, and it's at the end that we wrote the result into its decimal form (aka 1.25)

Now for m/n, we have been ask to prove that the result is "either zero repetend or repeating decimal", nothing more. We don't care to know how many decimal there is if it's repetend or how many decimal there are if repeating. For this we need to focus on the property of the leftover, and as you said above:

lets do the step one m=q1*n+r1
so r1 can only be equal to only one of those integers: 0; 1; 2; 3; .....; n-2; n-1, note that there is only (n-1) possible value that isn't 0 (earlier, in 5/4 leftover could only have the value 0; 1; 2; 3)
Now if r1=0 then we're done, if not we have to divide r1 by n for step2
r1=q2*n+r2
Now if r2=0 we are done, if r2=r1 then we have entered a cycle. Else we have to divide r2 by n again for step3
r2=q3*n+r3
if r3=0 we are done, if r3 equals r1 or r2 we have entered a cycle and are done, else I'll divide r3 by n for step4

and so on and so forth

What matter here is that since there is a limited number of possible value for the leftover, I'm bound either to finally get 0 or any leftover I've already found previously. 0≤ r <n with r and n being integer are the key to prove that m/n is either zero repetend or repeating decimal (technically we proved it can't be another way)
.

Chriscase said "the long division algorithm determines that the sequence of remainders will repeat"

Is this really the case? I think a perfect answer should include also this kind of statement. There was a written question, which hinted that repeating decimals could occur. But can you trust the author, in this case? Trust but verify I say.

Because long division is being used to prove something. Long division itself has been proven by axioms. I think a written statement should be made, to a written question "why".

But otherwise very nice answer! Thank you for elaborating, your procedure, I suppose that's just the same procedure as long division, simply written out in that kind of form? Also, what type of proof, would your proof constitute as?

5. ## Re: Need help with science/math schoolwork? Post here!

There seems to be a bit of a language barrier here. Town Watch, are you saying you understand the proof, or is there still something you are asking for clarification about?

6. ## Re: Need help with science/math schoolwork? Post here!

Originally Posted by John Doe
When you decompose the long division of 5 by 4, the first step is that 5= 1*4+1
then you multiply the leftover, here 1, by 10 to get the next decimal number and you start again: 1*10=10=2*4+2
here the leftover (here 2), it isn't 0 or the same as before (it's not 1 again) so you carry on the procedure by multiplying the leftover by 10 and dividing the result by 4: 2*10=20=5*4+0

This time the leftover is zero therefore you stop and you actually find 1.25, which is also 125/100.... Even though 1.25 isn't an integer, what matter is that at each step we took the leftover of each division was one, and it's at the end that we wrote the result into its decimal form (aka 1.25)

Now for m/n, we have been ask to prove that the result is "either zero repetend or repeating decimal", nothing more. We don't care to know how many decimal there is if it's repetend or how many decimal there are if repeating. For this we need to focus on the property of the leftover, and as you said above:

lets do the step one m=q1*n+r1
so r1 can only be equal to only one of those integers: 0; 1; 2; 3; .....; n-2; n-1, note that there is only (n-1) possible value that isn't 0 (earlier, in 5/4 leftover could only have the value 0; 1; 2; 3)
Now if r1=0 then we're done, if not we have to divide r1 by n for step2
r1=q2*n+r2
Now if r2=0 we are done, if r2=r1 then we have entered a cycle. Else we have to divide r2 by n again for step3
r2=q3*n+r3
if r3=0 we are done, if r3 equals r1 or r2 we have entered a cycle and are done, else I'll divide r3 by n for step4

and so on and so forth

What matter here is that since there is a limited number of possible value for the leftover, I'm bound either to finally get 0 or any leftover I've already found previously. 0≤ r <n with r and n being integer are the key to prove that m/n is either zero repetend or repeating decimal (technically we proved it can't be another way)
.

for 254178 / 7855

any leftover of any step could only have the value 0; 1; 2; 3;.....; 78853; 78854.
so if after 78854 steps I haven't found a leftover that is 0 or one that I have already found, then I will at step 78855 because there isn't any other option left.

because if the numbers are the same, so will the result of their division, what matter here is the value of ri, the value of qi is irrelevent.

because 0=<r<n, r and n being integers, it's easy to count how many number there are between 1 and n-1

I don't quite understand the notation, sorry

please calculate 1/17, in long division

define step1, step2, step3

show r1, r2, r3

show q1, q2, q3,

When I was taught long division at school it was simply rote memorization e.g.

-compare how many times divisor, fits into dividend's first digit(from the left, like in long division) Then you put zero. Add comma after the zero.

-Then you compare how many times divisor fits into dividend's two first digits. Then you put zero.

- Compare how many times divisor fits into dividend's three first digits. Then you put five, because five times 17 equals to 85 (85 smaller than 100)

- then you multiply 5 by 17.

-then you subtract 85, from 100= 15

-then you compare how many times does 17 fit into 15. You make a mental note zero times.

-then you drop a ten, and compare how many times does 17 fit into 150

etc...

Then the cycle goes on again... It becomes difficult to see the step1 step2 etc...

But I think that I understand better the logical foundation of your proof.

It was something like this, in my brain.

Why does your proof show, that it cannot be otherwise? Why cannot long division m/n = irrational number? The axiom of rational numbers must be enforced before we start long division. (I suppose it must be shown or demonstrated that irrational number is impossible to be the quotient of two integers)

Therefore we know that m/n = rational number always. Also m,n belong to domain of integers.

The basic characteristic of rational number is that any rational number in existance can be shown as quotient, or fraction of two integers m,n

also n is unequal to 0

From these assumptions, it follows that m/n is a simple fraction. Therefore (because of integers being domain) Simple Fraction is same as division.

The basic characteristic of simple fraction is that the divisor, must be "any one countable and finite integer", in existance, except zero. Divisor can be as large or small as possible (just like dividend can be as large or small as possible), but it is always "a countable finite integer" (because simple fraction)

Simple fraction m/n can be divided in long division. But, I'm not sure why... we must calculate in long division? What does the calculation in long divison actually prove? We can already say that assuming the axioms about rational numbers being true, then we can simply claim with truthfulness, that long division will not result in irrational number. Sorry for being pessimistic

7. ## Re: Need help with science/math schoolwork? Post here!

I made a little mistake in my previous post, here it's corrected in bold

lets do the step one m=q1*n+r1
so r1 can only be equal to only one of those integers: 0; 1; 2; 3; .....; n-2; n-1, note that there is only (n-1) possible value that isn't 0 (earlier, in 5/4 leftover could only have the value 0; 1; 2; 3)
Now if r1=0 then we're done, if not we have to divide 10*r1 by n for step2
10*r1=q2*n+r2
Now if r2=0 we are done, if r2=r1 then we have entered a cycle. Else we have to divide 10*r2 by n again for step3
10*r2=q3*n+r3
if r3=0 we are done, if r3 equals r1 or r2 we have entered a cycle and are done, else I'll divide 10*r3 by n for step4

and so on and so forth

It was something like this, in my brain.

Why does your proof show, that it cannot be otherwise? Why cannot long division m/n = irrational number? The axiom of rational numbers must be enforced before we start long division. (I suppose it must be shown or demonstrated that irrational number is impossible to be the quotient of two integers)

Therefore we know that m/n = rational number always. Also m,n belong to domain of integers.

The basic characteristic of rational number is that any rational number in existance can be shown as quotient, or fraction of two integers m,n

also n is unequal to 0

From these assumptions, it follows that m/n is a simple fraction. Therefore (because of integers being domain) Simple Fraction is same as division.

The basic characteristic of simple fraction is that the divisor, must be "any one countable and finite integer", in existance, except zero. Divisor can be as large or small as possible (just like dividend can be as large or small as possible), but it is always "a countable finite integer" (because simple fraction)

Simple fraction m/n can be divided in long division. But, I'm not sure why... we must calculate in long division? What does the calculation in long divison actually prove? We can already say that assuming the axioms about rational numbers being true, then we can simply claim with truthfulness, that long division will not result in irrational number. Sorry for being pessimistic
No, I did prove that m/n is either zero repetend or repeating decimal, because you have no other choice but to either find a leftover that is nul or to find a leftover that we have already found (start of the repeating decimal).

8. ## Re: Need help with science/math schoolwork? Post here!

Thank you for correcting for the mistake, I was originally wondering why the notation was like that, but I was too embarrased to ask about "why you do not account for multiplied by 10, it's standard procedure in long division"

9. ## Re: Need help with science/math schoolwork? Post here!

Originally Posted by John Doe
When you divide integers n by m, n=q*m+r where r is the leftover: you know its value is 0 or any integer strictly below m, basically r belong to {0,1,2,...,m-1} a finite set of numbers. If the leftover isn't 0, then you multiply the leftover by 10 and try to divide it again by m (to get the next decimal of you result), and you repeat the process until either the leftover is 0 or of the same value of a previous leftover (start of the cycle). So at the worst case scenario, if you havent got a leftover equal to zero or equal to a previous leftover at the (m-1)th division, you will at the mth division because you have exhausted all other possible result.

I did not demonstrate that m/n is either zero repetend or repeating decimal, I demonstrated it can't be otherwise (it's obviously not very rigourous but it will do).

1. I think I understand you proof now after having studied it for some time. I assume that your proof is combinatorial proof, on the basis of long division? Just like in deck of cards, where number of cards is 10, there exist only limited number of possible combinations of two cards etc...

2. From where exactly do you conclude that r must belong to finite set of natural numbers? when you divide integers m,n such that m/n
I have difficulty understanding where this condition comes from. Can the reasoning be explained in words preferrably? Do you claim that this is because of the inequality portion of long division algorithm?

Forgive me for not being good at this problem, I only learned about the mathematical formula behind long division algorithm two days ago.

10. ## Re: Need help with science/math schoolwork? Post here!

]
Originally Posted by Town Watch
1. I think I understand you proof now after having studied it for some time. I assume that your proof is combinatorial proof, on the basis of long division? Just like in deck of cards, where number of cards is 10, there exist only limited number of possible combinations of two cards etc...
yes, but to be precise in this case, it's how many possible combination of 1 card I can pick from a packet of n card, obviously it's n. if I were to pick a card (n+1) times, it's impossible not to find the same card twice because there's only n card (you put the card back into the deck and mix the deck before picking again). I don't care it's possible to pick the same card earlier, what I care is that the maximum of picks (worst case scenario) before I pick the same card twice is a finite number, it's impossible to always find a different card since we pick from the same deck each time.

Originally Posted by Town Watch
2. From where exactly do you conclude that r must belong to finite set of natural numbers? when you divide integers m,n such that m/n
because m=q*n+r with m,n,q and r being integers and 0=<r<n
so r can only take one of an integer value between 0 and n-1, they are {0;1;2;3;...;n-3;n-2;n-1} we know that it's a finite number because n is fixed, even though we don't know the value of n. We don't know either any value of any leftover from any step of the long division, but we know that each value is one within the set {0;1;2;3;....;n-2;n-1} since we always divide everything by the same number n.

if n=4, then the leftover can only be one of {0;1;2;3} or one of 4 possible value, I don't care which one. I know that after 4 steps in the long division, I already got the leftover 0 or one of the others {1;2;3}.
if n=7855 then the leftover can only be one of {0;1;2;3;.....;7853;7854} or one of 7855 possible value, I don't care which one. I know that after 7855 steps I already got the leftover 0 or one of the others {1;2;3;...7853;7854}.

Either one leftover will be zero and it end the division, or I step into a loop if I find a leftover I found earlier (I don't need to know how big the loop will be, I just need to know there's one)

For a division m/n not to have a zero repetend or a repeating decimal, I would need an infinite number of possible leftover, it's impossible because we always divide everything by n, and there is a fixed number of possible leftovers and they are {0;1;2;3;...;n-2;n-1}.

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