Union (u) and Intersection (n)

**mj.alawami** · January 15th 2009, 01:24 PM

the real number system
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eg. Q n R = Q

Find?
1)Q u H
2)J u N
3)H n J
4)H n R
5)Q u R
6)H n Q
7)J n N

I cant understand how to do them ? please solve the above with explanation.

**Last_Singularity** · January 15th 2009, 01:41 PM

Originally Posted by mj.alawami

the real number system
ImageShack - Image Hosting :: 16012009078os8.jpg

eg. Q n R = Q

Find?
1)Q u H
2)J u N
3)H n J
4)H n R
5)Q u R
6)H n Q
7)J n N

I cant understand how to do them ? please solve the above with explanation.

W, J, H? Abusive notation. Anyway, lol...

1) It tells you that the real numbers are composed of irrational and rationals. In other words, real numbers are composed of numbers that cannot be expressed as a fraction of integers (irrational ones) and those that can be expressed as a fraction of integers (rational ones). Thus:
$\mathbb{H} \cup \mathbb{Q} = \mathbb{R}$

2) Clearly the natural numbers are a subset of the integers. Any natural number in the set {1,2,3,...} is a part of the set of integers {...,-2,-1,0,1,2,3,...}. So their union is just the integers again. Thus:
$\mathbb{N} \cup \mathbb{J} = \mathbb{J}$

3) The intersection of the irrationals and the integers is the set of all elements that BOTH irrational and integer. Can you find any such numbers? Can you find numbers that cannot be expressed as a ratio between two integers and yet are still integers?

4) The irrationals a proper subset of the reals, so if you are asking the intersection, it is the set of all numbers that both real and irrational. But by definition, all irrationals are real. So the set that you get for your answer is just the irrationals once again.

In general, here are a few tips for sets:
-If $X$ is a subset of $Y$ (written as $X \subseteq Y$ ), then their intersection is $X \cap Y = Y$ . Example: the irrationals are a subset of the reals, so their intersection is just the irrationals.
-If $X$ is a subset of $Y$ , then their union is $X \cup Y = Y$ . Example: the irrationals are a subset of the reals, so their union is just the reals.
-If $X$ and $Y$ are disjoint sets (meaning that they share no elements), then their intersection is the empty set. Example: the intersection between the negative integers and the positives is empty (no such number is both negative and positive at the same time)
-If $X$ and $Y$ are disjoint sets, then their union is simply a new set that contains every element of both $X$ and $Y$ . Example: the union of all the nonnegative integers and the negative integer is simply the set of integers because every integer is either negative or nonnegative.

Try to finish the rest yourself and show us your work. We can check it over for you.

**mj.alawami** · January 15th 2009, 11:24 PM

Originally Posted by Last_Singularity

W, J, H? Abusive notation. Anyway, lol...

1) It tells you that the real numbers are composed of irrational and rationals. In other words, real numbers are composed of numbers that cannot be expressed as a fraction of integers (irrational ones) and those that can be expressed as a fraction of integers (rational ones). Thus:
$\mathbb{H} \cup \mathbb{Q} = \mathbb{R}$

2) Clearly the natural numbers are a subset of the integers. Any natural number in the set {1,2,3,...} is a part of the set of integers {...,-2,-1,0,1,2,3,...}. So their union is just the integers again. Thus:
$\mathbb{N} \cup \mathbb{J} = \mathbb{J}$

3) The intersection of the irrationals and the integers is the set of all elements that BOTH irrational and integer. Can you find any such numbers? Can you find numbers that cannot be expressed as a ratio between two integers and yet are still integers?

4) The irrationals a proper subset of the reals, so if you are asking the intersection, it is the set of all numbers that both real and irrational. But by definition, all irrationals are real. So the set that you get for your answer is just the irrationals once again.

In general, here are a few tips for sets:
-If $X$ is a subset of $Y$ (written as $X \subseteq Y$ ), then their intersection is $X \cup Y = Y$ . Example: the irrationals are a subset of the reals, so their intersection is just the irrationals.
-If $X$ is a subset of $Y$ , then their union is $X \cap Y = Y$ . Example: the irrationals are a subset of the reals, so their union is just the reals.
-If $X$ and $Y$ are disjoint sets (meaning that they share no elements), then their intersection is the empty set. Example: the intersection between the negative integers and the positives is empty (no such number is both negative and positive at the same time)
-If $X$ and $Y$ are disjoint sets, then their union is simply a new set that contains every element of both $X$ and $Y$ . Example: the union of all the nonnegative integers and the negative integer is simply the set of integers because every integer is either negative or nonnegative.

Try to finish the rest yourself and show us your work. We can check it over for you.

My attempt to answer -
3) H n J=R
4)H n R=H
5)Q u R=R
6)H n Q=R
7)J n N =N

If they any one of them are wrong please tell me which on and explain it ,Thank you

**Aryth** · January 16th 2009, 05:13 AM

3) You're saying that the irrational numbers and the integers, when intersected, create the real numbers?

Think about it this way... Irrational numbers are the numbers that have no fractional form, and can therefore NEVER be integers. And the set of Integers is the set of all WHOLE numbers, and can therefore never be IRRATIONAL. It is clear that these have nothing in common and are therefore disjoint sets, meaning:

$\mathbb{H} \cap \mathbb{J} = \emptyset$

4) That one is right.

5) That one is also right.

6) I see that your achilles heel is with disjoint sets.

Think about this one, The set of the irrational numbers are numbers that have no fractional form, and can therefore NEVER be rational. And the set of rational numbers is the set of numbers that CAN be put into fractional form, and can therefore NEVER be irrational. It is clear that these have nothing in common and are therefore disjoint sets, meaning:

$\mathbb{H} \cap \mathbb{Q} = \emptyset$

7) That is right.

**mj.alawami** · January 16th 2009, 09:26 PM

Originally Posted by Aryth

3) You're saying that the irrational numbers and the integers, when intersected, create the real numbers?

Think about it this way... Irrational numbers are the numbers that have no fractional form, and can therefore NEVER be integers. And the set of Integers is the set of all WHOLE numbers, and can therefore never be IRRATIONAL. It is clear that these have nothing in common and are therefore disjoint sets, meaning:

$\mathbb{H} \cap \mathbb{J} = \emptyset$

4) That one is right.

5) That one is also right.

6) I see that your achilles heel is with disjoint sets.

Think about this one, The set of the irrational numbers are numbers that have no fractional form, and can therefore NEVER be rational. And the set of rational numbers is the set of numbers that CAN be put into fractional form, and can therefore NEVER be irrational. It is clear that these have nothing in common and are therefore disjoint sets, meaning:

$\mathbb{H} \cap \mathbb{Q} = \emptyset$

7) That is right.

Can you please check these question too for me ...
N u Q =Q
J n N =N
Q n H = /0 (disjoint)
Q u J =Q

**Last_Singularity** · January 17th 2009, 04:37 AM

Originally Posted by mj.alawami

Can you please check these question too for me ...
N u Q =Q
J n N =N
Q n H = /0 (disjoint)
Q u J =Q

That is correct - good job!

**mj.alawami** · January 17th 2009, 06:39 AM

Originally Posted by Last_Singularity

That is correct - good job!

Thank you very much ... Can't put the rest in words

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