is there something wrong in this question? "Show that if a, b are positive integers, then there is smallest positive integer a−bk, k an integer"

[logistic_guy]

Show that if $\displaystyle a$ and $\displaystyle b$ are positive integers, then there is a smallest positive integer of the form $\displaystyle a - bk, k \in \bold{Z}$.

 

[Dr.Peterson]

Show that if $\displaystyle a$ and $\displaystyle b$ are positive integers, then there is a smallest positive integer of the form $\displaystyle a - bk, k \in \bold{Z}$.

Tell us your thoughts! Is there something you think is wrong with it? Does it seem too easy, or too hard, or impossible, or what?

And I assume you've done what I'd do in your place, just pondering an example. Is there a smallest positive integer of the form, say, 2 - 3k, where k is an integer? What does that mean? What is that number?

 

[logistic_guy]

i think the question wrong because when $\displaystyle k \geq 0$, the expression isn't true

look at your example

2 - 3k,

$\displaystyle k = 1$ make it fail

 

[Otis]

when $\displaystyle k \geq 0$, the expression isn't true

$\displaystyle k = 1$ make it fail

Hi. When you say, "the expression isn't true", I think you mean, "a–bk is not positive".

The exercise does not claim that a–bk is positive for all values of k, so it's okay if some k-values make a–bk negative.

Other k-values will make a–bk positive. Of all possible, positive values of a–bk, the exercise asks which one is smallest requires you to show that there is a lower bound.

 

[Dr.Peterson]

i think the question wrong because when $\displaystyle k \geq 0$, the expression isn't true

look at your example

$\displaystyle k = 1$ make it fail

So your answer to my first question is that you think it's impossible?

Try actually answering the questions I asked, which were intended to redirect your thinking away from the direction I thought you might be going:

Is there a smallest positive integer of the form, say, 2 - 3k, where k is an integer? What does that mean? What is that number?

And don't forget that Z includes both positive and negative integers.

But on the other hand, don't forget that the problem doesn't actually ask you to find the number; it asks you to prove that it always exists:

Show that if $\displaystyle a$ and $\displaystyle b$ are positive integers, then there is a smallest positive integer of the form $\displaystyle a - bk, k \in \bold{Z}$.

There is a property of the positive integers that makes this easy to prove, even if it were impossible to actually find such a number.

 

[logistic_guy]

Hi. When you say, "the expression isn't true", I think you mean, "a–bk is not positive".

The exercise does not claim that a–bk is positive for all values of k, so it's okay if some k-values make a–bk negative.

Other k-values will make a–bk positive. Of all possible, positive values of a–bk, the exercise asks which one is smallest requires you to show that there is a lower bound.

thank you Otis, i understood the question wrong. i thought it must true for all values of $\displaystyle k$

And don't forget that Z includes both positive and negative integers.

i know what $\displaystyle \bold{Z}$ is, i just interpretd the question wrong thinking i have to show for all $\displaystyle k$ it is valid.

So your answer to my first question is that you think it's impossible?

Try actually answering the questions I asked, which were intended to redirect your thinking away from the direction I thought you might be going:

And don't forget that Z includes both positive and negative integers.

But on the other hand, don't forget that the problem doesn't actually ask you to find the number; it asks you to prove that it always exists:

There is a property of the positive integers that makes this easy to prove, even if it were impossible to actually find such a number.

i know we get positive integer when $\displaystyle k \leq 0$

my analysis is like this, $\displaystyle k = 0$ will destroy the form $\displaystyle a - bk$ to $\displaystyle a$ which is not the required form

for negative numbers, we keep the form safe $\displaystyle a - b(k)$, staring with $\displaystyle k = -1$, then $\displaystyle k = -2$, give us $\displaystyle a + b$ and $\displaystyle a + 2b$

since $\displaystyle a$ and $\displaystyle b$ are positive integers, the number must ascending. this means $\displaystyle a + b$ is smallest positive number

There is a property of the positive integers that makes this easy to prove, even if it were impossible to actually find such a number.

i'm sure you mean the Well-ordering property. the first part of the property is confusing me because it say for example, show the set not impty, of course the set isn't impty without showing that. the second part to find the smallest number is a little diffciult in some cases but in ours not.

 

[Dr.Peterson]

my analysis is like this, $\displaystyle k = 0$ will destroy the form $\displaystyle a - bk$ to $\displaystyle a$ which is not the required form

I wouldn't say it that way; it is still the same form even when a term becomes zero. Similarly, we can say that 2i is written in the form a+bi though a is zero.

Taking my example of 2 - 3k, for k = -2, -1, 0, 1, 2, the values are 8, 5, 2, -1, -4; all of these are considered to have the form a - bk, meaning that they can be expressed that way for some integer k.

for negative numbers, we keep the form safe $\displaystyle a - b(k)$, staring with $\displaystyle k = -1$, then $\displaystyle k = -2$, give us $\displaystyle a + b$ and $\displaystyle a + 2b$

since $\displaystyle a$ and $\displaystyle b$ are positive integers, the number must ascending. this means $\displaystyle a + b$ is smallest positive number

Not always. In my example, 2 - 3k, the smallest such number is 2, not 5 (because, again, k is allowed to be zero).

Take another example: When a=3 and b=2, the smallest positive value of a-bk is 3-2(1) = 1 (that is, a-b, not a+b).

Be careful of hasty generalization. (That's why proofs are needed.)

i'm sure you mean the Well-ordering property. the first part of the property is confusing me because it say for example, show the set not empty, of course the set isn't empty without showing that. the second part to find the smallest number is a little difficult in some cases but in ours not.

Please state the property as you learned it, so we can be sure what "the first part" and "the second part" refer to.

I would say the first step is to identify what set you are applying the property to; and the property doesn't require finding the smallest number at all! Without knowing what set you are talking about, it's impossible to show it is non-empty. When you do know, it's often easy.

 

[logistic_guy]

thank yu Dr.Peterson. nice explanation

let me try again to prove it

this form $\displaystyle a - bk$ give negative and postive integers, but we don't care for negative integrs

when $\displaystyle k = 0$, we can have the positive integers $\displaystyle 1,2,3,4,5,.......$

these positive integrs was produced from the form $\displaystyle a - bk$

let $\displaystyle A$ be a set contains these positive integers, then $\displaystyle A = \{1,2,3,4,5,.........\}$

i have this property A nonempty set of positive integers is said to satisfy the well-ordering property if it has the least element.

set $\displaystyle A$ is not empty because it contains the element $\displaystyle 1$

the smallest element in the set $\displaystyle A$ is $\displaystyle 1$, so the form $\displaystyle a - bk$ indeed have a smallest positive integer

 

[Dr.Peterson]

let me try again to prove it

What you're trying to prove, I assume, is what you said at the start:

Show that if $\displaystyle a$ and $\displaystyle b$ are positive integers, then there is a smallest positive integer of the form $\displaystyle a - bk, k \in \bold{Z}$.

So you need to prove existence of a particular integer.

this form $\displaystyle a - bk$ give negative and postive integers, but we don't care for negative integrs

when $\displaystyle k = 0$, we can have the positive integers $\displaystyle 1,2,3,4,5,.......$

these positive integrs was produced from the form $\displaystyle a - bk$

First, why do you think it matters what happens for k=0? It's true that in that case, a - bk is a, which is a positive integer; but I don't think that contributes anything to a proof of the general fact you are asked to prove.

let $\displaystyle A$ be a set contains these positive integers, then $\displaystyle A = \{1,2,3,4,5,.........\}$

This set obviously has a smallest element, 1; but how does that help?

i have this property A nonempty set of positive integers is said to satisfy the well-ordering property if it has the least element.

This is not my understanding of the well-ordering property; but I see a variety of statements about such a "property" or "principle". Can you show the source of your statement of this "property" (and also the source of your problem) so we can be sure it is appropriate to your context?

The well-ordering principle I find applicable is a property, not of some particular set of integers, but of the set of all positive integers:

In mathematics, the well-ordering principle states that every non-empty subset of positive integers contains a least element.​

Do you see the difference from what you wrote? Details matter!

I find the same statement, with the same name, here; that also generalizes the concept, saying

A set T of real numbers is said to be well-ordered if every nonempty subset of T has a smallest element.​

​

Therefore, according to the principle of well-ordering, N is well-ordered.​

So, this is a property that N has, and other sets may have, but not all sets do.

But in searching for a "well-ordering property", I find fewer references; here is one:

The well-ordering property of N states that “For all sets S ⊆ N such that S 6 = ∅, there exists a least element m ∈ S such that m ≤ t for all t ∈ S.”​

This says the same thing. Again, it is not as simple as what you wrote.

Again, do you see the difference? And do you see how the property/principle I am referring to leads to the proof you want?

set $\displaystyle A$ is not empty because it contains the element $\displaystyle 1$

the smallest element in the set $\displaystyle A$ is $\displaystyle 1$, so the form $\displaystyle a - bk$ indeed have a smallest positive integer

What you say here is irrelevant to what you want to prove. What subset of N is relevant?

 

[logistic_guy]

First, why do you think it matters what happens for k=0? It's true that in that case, a - bk is a, which is a positive integer; but I don't think that contributes anything to a proof of the general fact you are asked to prove.

you're correct Dr.Peterson. it doesn't matter for $\displaystyle k = 0$ and i should prove it in general

This is not my understanding of the well-ordering property; but I see a variety of statements about such a "property" or "principle". Can you show the source of your statement of this "property" (and also the source of your problem) so we can be sure it is appropriate to your context?

the property, i copy it from google, but i can't find it now. i take the question from brainly.com

i give up to solve this question because i gave everything i know. i'll read more about the well-ordering property definitions you give me and i'll try to combine it with a general proof

 

[Dr.Peterson]

you're correct Dr.Peterson. it doesn't matter for $\displaystyle k = 0$ and i should prove it in general


the property, i copy it from google, but i can't find it now. i take the question from brainly.com

i give up to solve this question because i gave everything i know. i'll read more about the well-ordering property definitions you give me and i'll try to combine it with a general proof

Here's a bigger hint: The subset of the natural numbers you need to consider is the set of all positive integers given by a - bk, for integers k.

I see the question on Brainly; you are right not to trust the answer given there.

 

[Steven G]

This is similar to long division where we try to compute a/b.
You start with a and keep subtracting b until you get 0 or the first negative value. Once you get zero or the first negative value you back off one.

 

[logistic_guy]

This is similar to long division where we try to compute a/b.
You start with a and keep subtracting b until you get 0 or the first negative value. Once you get zero or the first negative value you back off one.

you mean $\displaystyle a$ will be fixed. the question say it is positive integer. so i'll never be fixed

Here's a bigger hint: The subset of the natural numbers you need to consider is the set of all positive integers given by a - bk, for integers k.

I see the question on Brainly; you are right not to trust the answer given there.

let us assume i've set contain all positive integers produces by $\displaystyle a - bk$

does this guarantee this set won't have negative element? if $\displaystyle a$ positive and $\displaystyle b$ positive and $\displaystyle k$ positive and $\displaystyle b > a$, the form $\displaystyle a - bk$ produces a negative element

 

[Dr.Peterson]

you mean $\displaystyle a$ will be fixed. the question say it is positive integer. so i'll never be fixed

I'm guessing you meant "it'll never be fixed", not that you yourself are irreparable (or refuse to be spayed).

But saying that $a$ is a positive integer does not mean it can't be a fixed number.

The claim is that, given any values of $a$ and $b$ (which will then be fixed), we can consider the set of all numbers of the form $a-bk$ for those particular values of $a$ and $b$, and any integer $k$. That is, the values of $a$ and $b$ are fixed while we construct the set.

Why are you objecting to this? I think his point is a very good one. The remainder on division is exactly the smallest positive integer of this type.

let us assume i've [made the] set contain[ing] all positive integers produce[d] by $\displaystyle a - bk$

does this guarantee this set won't have [a] negative element? if $\displaystyle a$ positive and $\displaystyle b$ positive and $\displaystyle k$ positive and $\displaystyle b > a$, the form $\displaystyle a - bk$ produces a negative element

Actually, this set is guaranteed to contain both positive and negative elements (since b is not zero). Perhaps you are misreading what I said:

The subset of the natural numbers you need to consider is the set of all positive integers given by a - bk, for integers k.

That does not mean that the set $\{a-bk | k \in\mathbb{Z}\}$ is a set of positive integers; it means that we consider only the elements of $\{a-bk | k \in\mathbb{Z}\}$ that are positive; that is, I am talking about the intersection of that set with $\mathbb{Z}$.

Does that make sense?

 

[logistic_guy]

i think i'm understanding, but not fully. why i need a set contain both negative and positive integers? why not making a set contain only positive integers directly?

 

[Dr.Peterson]

i think i'm understanding, but not fully. why i need a set [that contains] both negative and positive integers? why not [make] a set [that contains] only positive integers directly?

Tell me how you would do that!

The problem itself provides a set; that set is a set of integers; to solve the problem, you have to find the smallest positive integer in it.

You use what you are given, and make what you need to make.

 

[logistic_guy]

so i have to have the set of natural numbers. it have the subset of positive integers. then i can use the well-ordering property because i'm dealing with a subset. correct?

 

[Dr.Peterson]

so i have to have the set of natural numbers. it have the subset of positive integers. then i can use the well-ordering property because i'm dealing with a subset. correct?

I can't tell exactly what you are saying.

The natural numbers are either the positive integers, or the non-negative integers, depending on your definition. But are you talking about the natural numbers being a subset of the positive integers (are they?), or about another set, which we have discussed, being a subset of the positive integers? Your use of the word "the" is confusing. (It makes a big difference in English.)

 

[Steven G]

Please solve a-bk>0 for k and then finish the proof.

 

[logistic_guy]

I can't tell exactly what you are saying.

The natural numbers are either the positive integers, or the non-negative integers, depending on your definition. But are you talking about the natural numbers being a subset of the positive integers (are they?), or about another set, which we have discussed, being a subset of the positive integers? Your use of the word "the" is confusing. (It makes a big difference in English.)

let me rephrase what i mean with symbols and tell me where i'm wrong

the well-ordering property is apply only to a subset

if $\displaystyle N$ is the set of natural numbers and $\displaystyle P$ is the set of positive integers with form $\displaystyle a - bk$

then $\displaystyle P$ is a subset of $\displaystyle N$

since $\displaystyle P$ is a subset now, the well-ordering property can be apply

Please solve a-bk>0 for k and then finish the proof.

easy
$\displaystyle a-bk>0$
$\displaystyle a>bk$
$\displaystyle \frac{a}{b}>k$

i'm trying to finish the proof with the the well-ordering property