Elements of the set {6a + 2b} where a and b....

[Math100]

Homework Statement

Write each of the following sets by listing their elements between braces.

Relevant Equations

None.

I've boxed around my answer but I don't know if this is the right answer. I also used a substitution with c=3a+b in my work. I don't know if my work is precise or not.

 

[fresh_42]

Yes, it's correct. Do you see what $a\cdot\mathbb{Z} + b\cdot\mathbb{Z}$ is in general?

 

[Math100]

Yes, it's correct. Do you see what $a\cdot\mathbb{Z} + b\cdot\mathbb{Z}$ is in general?

It's ℤ(a+b).

 

[Math100]

It's ℤ(a+b).

Where ℤ represents a set of integers {...-3, -2, -1, 0, 1, 2, 3...} and I multiplied this set of integers by 2 to get the answer/result.

 

[fresh_42]

It's ℤ(a+b).

No. You have shown that $6\mathbb{Z}+2\mathbb{Z}=2\mathbb{Z}$.

$\mathbb{Z}(a+b)$ would have been $8\mathbb{Z}$ which is less. If you gather all multiples of $a$ and add all multiples of $b$, then you get all multiples of?

 

[Math100]

No. You have shown that $6\mathbb{Z}+2\mathbb{Z}=2\mathbb{Z}$.

$\mathbb{Z}(a+b)$ would have been $8\mathbb{Z}$ which is less. If you gather all multiples of $a$ and add all multiples of $b$, then you get all multiples of?

8ℤ

 

[fresh_42]

My question is: which numbers can you build of the form $\{n=x\cdot a+y\cdot b\,|\,x,y\in \mathbb{Z}\}$?

You had $a=6$ and $b=2$ and found all even numbers as the solution: $6\mathbb{Z}+2\mathbb{Z}=2\mathbb{Z}.$

If you take $a=6$ and $b=3$, what do you get? And what for $a=6$ and $b=5$?

 

[Math100]

I get multiples of b.

 

[fresh_42]

Yes, but if you add multiples of $a$ and multiples of $b$, then you can get more. See
$6\mathbb{Z}+2\mathbb{Z}=2\mathbb{Z}$ - your exercise, which was correct.

If we calculate $6\mathbb{Z}+3\mathbb{Z}$ then we get all multiples of three, $3\mathbb{Z}$, because the multiples of $6$ are already included.

But if we calculate $6\mathbb{Z}+5\mathbb{Z}$ then we have $1\cdot 6+ (-1)\cdot 5=1$ in our set. But all multiples of $1$ are all integers, so $6\mathbb{Z}+5\mathbb{Z}=1\cdot \mathbb{Z}=\mathbb{Z}.$

 

[Math100]

My question is: which numbers can you build of the form $\{n=x\cdot a+y\cdot b\,|\,x,y\in \mathbb{Z}\}$?

You had $a=6$ and $b=2$ and found all even numbers as the solution: $6\mathbb{Z}+2\mathbb{Z}=2\mathbb{Z}.$

If you take $a=6$ and $b=3$, what do you get? And what for $a=6$ and $b=5$?

If I take a=6 and b=3, then I get 6ℤ+3ℤ=3ℤ where the set is {...,-9, -6, -3, 0, 3, 6, 9,...}.
If I take a=6 and b=5, then I get 6ℤ+5ℤ=5ℤ where the set is {...,-15, -10, -5, 0, 5, 10, 15,...}.

 

[fresh_42]

If I take a=6 and b=3, then I get 6ℤ+3ℤ=3ℤ where the set is {...,-9, -6, -3, 0, 3, 6, 9,...}.
If I take a=6 and b=5, then I get 6ℤ+5ℤ=5ℤ where the set is {...,-15, -10, -5, 0, 5, 10, 15,...}.

This wasn't completely wrong. The first line is correct. Only the second does not have enough numbers in the set.

We always get multiples of $\mathbb{Z}$, in all cases. The question is: which multiples?

 

[Math100]

Yes, but if you add multiples of $a$ and multiples of $b$, then you can get more. See
$6\mathbb{Z}+2\mathbb{Z}=2\mathbb{Z}$ - your exercise, which was correct.

If we calculate $6\mathbb{Z}+3\mathbb{Z}$ then we get all multiples of three, $3\mathbb{Z}$, because the multiples of $6$ are already included.

But if we calculate $6\mathbb{Z}+5\mathbb{Z}$ then we have $1\cdot 6+ (-1)\cdot 5=1$ in our set. But all multiples of $1$ are all integers, so $6\mathbb{Z}+5\mathbb{Z}=1\cdot \mathbb{Z}=\mathbb{Z}.$

I didn't know that if I take a=6 and b=5, I get that result. This is similar to modulo mathematics problems.

 

[Math100]

{5a+2b: a, b ∈ ℤ}
{c=5a+2b ∈ ℤ}
{c ∈ ℤ}
{...-3, -2, -1, 0, 1, 2, 3,...}
Is the work precise? I took a=5 and b=2 for another practice problem.

 

[fresh_42]

{5a+2b: a, b ∈ ℤ}
{c=5a+2b ∈ ℤ}
{c ∈ ℤ}
{...-3, -2, -1, 0, 1, 2, 3,...}
Is the work precise? I took a=5 and b=2 for another practice problem.

Yes, that is true. It is all about which is the smallest number we can build in the set.

 

[Math100]

You know, that some people use q for the substitution instead of c during the work process. I used c=5a+2b. I don't know which is the accurate/precise notation.

 

[fresh_42]

You know, that some people use q for the substitution instead of c during the work process. I used c=5a+2b. I don't know which is the accurate/precise notation.

That doesn't matter. If it comes to integers, then often $p,q,r,s,t$ are used, where $p$ and $q$ are reserved for primes. But this isn't a rule, just a convention. If physics comes into play, then $s$ (path) and $t$ (time) aren't available. I had a professor who liked to use $\mathfrak{a,b,c,d,e,f,g,h,i,j,k,l,m,n,}$ etc.

 

[Math100]

That doesn't matter. If it comes to integers, then often $p,q,r,s,t$ are used, where $p$ and $q$ are reserved for primes. But this isn't a rule, just a convention. If physics comes into play, then $s$ (path) and $t$ (time) aren't available. I had a professor who liked to use $\mathfrak{a,b,c,d,e,f,g,h,i,j,k,l,m,n,}$ etc.

Thank you for the help!

 

[FactChecker]

Homework Statement:: Write each of the following sets by listing their elements between braces.
Relevant Equations:: None.

I've boxed around my answer but I don't know if this is the right answer. I also used a substitution with c=3a+b in my work. I don't know if my work is precise or not.

If I take your work literally, you got the right answer for the wrong reason.
When you factored out the 2 in the third line, you dropped a very important |a,b $\in \mathbb {Z}$.
That is not the same as what you have: $3a+b\in \mathbb {Z}$ with nothing said about what values a and b can have.
With |a,b $\in \mathbb {Z}$, suppose, you have a=0. Then you get the full set of values in the answer just by letting b take on all values in $\mathbb {Z}$.

 

[Math100]

If I take your work literally, you got the right answer for the wrong reason.
When you factored out the 2 in the third line, you dropped a very important |a,b $\in \mathbb {Z}$.
That is not the same as what you have: $3a+b\in \mathbb {Z}$ with nothing said about what values a and b can have.
With |a,b $\in \mathbb {Z}$, suppose, you have a=0. Then you get the full set of values in the answer just by letting b take on all values in $\mathbb {Z}$.

 

[Math100]

How about this time? Is the work right?

 

[fresh_42]

How about this time? Is the work right?

The point is: Why is $\{3\cdot a+1\cdot b\,|\,a,b\in \mathbb{Z}\}=\{q\in \mathbb{Z}\}=\mathbb{Z}$? This was what my questions were meant for. We always have $a\mathbb{Z}+b\mathbb{Z}=d\mathbb{Z}$. But what is $d$ in terms of $a$ and $b$?

Hint: $d$ is not the minimum! Consider $a=4$ and $b=6$.

 

[Math100]

The point is: Why is $\{3\cdot a+1\cdot b\,|\,a,b\in \mathbb{Z}\}=\{q\in \mathbb{Z}\}=\mathbb{Z}$? This was what my questions were meant for. We always have $a\mathbb{Z}+b\mathbb{Z}=d\mathbb{Z}$. But what is $d$ in terms of $a$ and $b$?

Hint: $d$ is not the minimum! Consider $a=4$ and $b=6$.

Because q=3a+b.

 

[fresh_42]

Because q=3a+b.

No. This is just a definition.

We have $\underbrace{\{\ldots,-6,-3,0,3,6,\ldots\}}_{3a}\, +\,\underbrace{\{\ldots,-2,-1,0,1,2,\ldots\}}_{b}$ so by setting $a=0$ we already have all integers via the second term, which means that all integers are included in the sum. The other direction, that the sum is included in the integers is by definition obvious.

What is $\{4a+6b\,|\,a,b\in \mathbb{Z}\}$?

All those sets can be written as $d\cdot \{\ldots,-2,-1,0,1,2,\ldots\}= \{\ldots,-2d,-d,0,d,2d,\ldots\}$, and $d$ is a function of $a$ and $b\; , \;d=f(a,b)$. Which function is $f$? $f$ is neither the maximum of $a$ and $b$ nor the minimum of $a$ and $b$, but what is it?

 

[Math100]

No. This is just a definition.

We have $\underbrace{\{\ldots,-6,-3,0,3,6,\ldots\}}_{3a}\, +\,\underbrace{\{\ldots,-2,-1,0,1,2,\ldots\}}_{b}$ so by setting $a=0$ we already have all integers via the second term, which means that all integers are included in the sum. The other direction, that the sum is included in the integers is by definition obvious.

What is $\{4a+6b\,|\,a,b\in \mathbb{Z}\}$?

All those sets can be written as $d\cdot \{\ldots,-2,-1,0,1,2,\ldots\}= \{\ldots,-2d,-d,0,d,2d,\ldots\}$, and $d$ is a function of $a$ and $b\; , \;d=f(a,b)$. Which function is $f$? $f$ is neither the maximum of $a$ and $b$ nor the minimum of $a$ and $b$, but what is it?

{4a+6b: a, b ∈ℤ}
{4a+6b \vert a, b ∈ℤ}
=2*{2a+3b \vert a, b∈ℤ}
=2*{q=2a+3b \vert a, b∈ℤ}
=2*{q∈ℤ}
={..., -6, -4, -2, 0, 2, 4, 6, ...}

 

[fresh_42]

{4a+6b: a, b ∈ℤ}
{4a+6b \vert a, b ∈ℤ}
=2*{2a+3b \vert a, b∈ℤ}
=2*{q=2a+3b \vert a, b∈ℤ}
=2*{q∈ℤ}
={..., -6, -4, -2, 0, 2, 4, 6, ...}

Yes, correct. $4\mathbb{Z}+6\mathbb{Z}=2\mathbb{Z}$. And we already know that $6\mathbb{Z}+2\mathbb{Z}=2\mathbb{Z}$ and that $3\mathbb{Z}+1\mathbb{Z}=1\mathbb{Z}$.

So we always have $a\mathbb{Z}+b\mathbb{Z}=d\mathbb{Z}$.
Our examples showed: $d(4,6)=2\, , \,d(3,1)=1\, , \,d(6,2)=2$.
Do you have a guess what $d$ is?

 

[Math100]

Yes, correct. $4\mathbb{Z}+6\mathbb{Z}=2\mathbb{Z}$. And we already know that $6\mathbb{Z}+2\mathbb{Z}=2\mathbb{Z}$ and that $3\mathbb{Z}+1\mathbb{Z}=1\mathbb{Z}$.

So we always have $a\mathbb{Z}+b\mathbb{Z}=d\mathbb{Z}$.
Our examples showed: $d(4,6)=2\, , \,d(3,1)=1\, , \,d(6,2)=2$.
Do you have a guess what $d$ is?

d(a, b)
d=b modulo a

 

[fresh_42]

d(a, b)
d=b modulo a

This would depend on what is $a$ and what is $b$. E.g. $6 \mod 4 = 2$ and $4\mod 6 = 4.$

No, it is symmetric in $a$ and $b$. It has to do with the divisors of the pair.

 

[Math100]

This would depend on what is $a$ and what is $b$. E.g. $6 \mod 4 = 2$ and $4\mod 6 = 4.$

No, it is symmetric in $a$ and $b$. It has to do with the divisors of the pair.

Where d is the GCF of a and b.

 

[fresh_42]

Where d is the GCF of a and b.

Greatest common divisor is the usual term, I think.

If you consider the numbers which are in both sets, i.e. $a\mathbb{Z}\cap b\mathbb{Z}=m\mathbb{Z}$ then $m$ is the least common multiple.

 

[Math100]

Greatest common divisor is the usual term, I think.

If you consider the numbers which are in both sets, i.e. $a\mathbb{Z}\cap b\mathbb{Z}=m\mathbb{Z}$ then $m$ is the least common multiple.

Thank you for the help!

 

[PeroK]

I just want to say I would do this entirely differently. First, $6a +2b$ is an even integer. So, the set is a subset of the even integers. To show that it is all even integers, we take any even integer $2k$ and this is in our set with $a = 0$ and $b = k$.

Is that not the obvious approach here?

 

[fresh_42]

Is that not the obvious approach here?

This depends on whether you want to emphasize on:
$$a\mathbb{Z}+b\mathbb{Z} \subseteq d\mathbb{Z} \subseteq a\mathbb{Z}+b\mathbb{Z} \Longrightarrow a\mathbb{Z}+b\mathbb{Z} =d\mathbb{Z}$$
or on the fact that $\mathbb{Z}$ is a principle ideal domain with
$$
a\mathbb{Z}+b\mathbb{Z} =\operatorname{gcd}(a,b)\mathbb{Z} \quad\text{ and }\quad a\mathbb{Z}\cap b\mathbb{Z} =\operatorname{lcm}(a,b)\mathbb{Z}
$$

However, as always in real life, the truth originates in a completely different post. It was an example of how $\{2x\in \mathbb{Z}\,|\,|x|\leq 5\}$ could be written better than it was in an OP. @Math100 took this as a template to write equations of sets. Your approach needs a different template to make it written properly because it partly specifies the coefficients which are arbitrary at the beginning. I tried to avoid confusion and pointed out the gcd, not the equality of sets.