Discrete M: Show that if A ⊆ B and C ⊆ D, then A X C ⊆ B X D

[leo255]

Homework Statement

[/B]
Sorry that I wasn't able to fit everything in the title. I got 2/3 on this on my quiz, and am wondering what I did wrong, or could have done better. Thanks in advance.

Show that if A ⊆ B and C ⊆ D, then A X C ⊆ B X D

Homework Equations

The Attempt at a Solution

For a given value x, if A ⊆ B, then x ∈ A and x ∈ B.
For a given value x, if C ⊆ D, then x i∈ C and x ∈ D.

Cartesian product of (A, C) means that all ordered pairs, (a, c) are included.
Cartesian product of (B, D) means that all ordered pairs, (b, d) are included.

A X C ⊆ B X D

 

[sxal96]

Do you know why your instructor docked you points on your quiz?

Here's an alternative route to getting started that's based on what you already have:
For some x ∈ A ⊆ B, then x ∈ A and x ∈ B. Similarly, for some y ∈ C ⊆ D, then y ∈ C and y ∈ D.

Cartesian products don't necessarily comprise of (x,x); we have to assume that there are two arbitrary elements of the two products, hence why I used (x,y).. We assume the statement above is true based off of what you are given to believe is true, which is that A ⊆ B and C ⊆ D. What conclusion can you draw from what we just stated in the italics?

 

[HallsofIvy]

You really haven't proven anything! You start by stating some definitions (always a good start) then simply assert the conclusion.

To prove "X\subset Y" start with "if p\in X" and use the definitions of X and Y to conclude "therefore p \in Y". Here X= A\times C. Now, if p\in A\times C, what can you say about p?

 

[LCKurtz]

Homework Statement

[/B]
Sorry that I wasn't able to fit everything in the title. I got 2/3 on this on my quiz, and am wondering what I did wrong, or could have done better. Thanks in advance.

Show that if A ⊆ B and C ⊆ D, then A X C ⊆ B X D

Homework Equations

The Attempt at a Solution

For a given value x, if A ⊆ B, then x ∈ A and x ∈ B.

That isn't the definition of A ⊆ B. Never mind that the statement isn't even true. You might start by looking up the correct definition of A ⊆ B.

 

[leo255]

That isn't the definition of A ⊆ B. Never mind that the statement isn't even true. You might start by looking up the correct definition of A ⊆ B.

Yes, you are correct - Math is not my strongest area, and I did not state that correctly. A being a subset means that A is a part of B (i.e. it is contained in B). Also, it is a proper subset if it is not equal to B.

 

[LCKurtz]

Yes, you are correct - Math is not my strongest area, and I did not state that correctly. A being a subset means that A is a part of B (i.e. it is contained in B). Also, it is a proper subset if it is not equal to B.

While that is an informal understanding, it is not the definition, and you need to use the correct definition to prove your proposition. The statement that A is a subset of B means if $a \in A$ then $a \in B$. So for your problem, you need to show, step by step, using what you are given, that if $p \in A\times C$ then $p \in B\times D$.

 

[HallsofIvy]

For a given value x, if A ⊆ B, then x ∈ A and x ∈ B.

If x\in A then x\in B

For a given value x, if C ⊆ D, then x i∈ C and x ∈ D.

If x\in c then x\in D