Advanced Calc. proof, about sets and intersection.

emira
Messages
7
Reaction score
0

Homework Statement


Prove that: A is a subset of B if and only if (A intersection B)=A


Homework Equations





The Attempt at a Solution

I tried proving the right side, that is

(A \cap B)=A
For two sets to be equal then they have to be subsets of each other...so:

(A \cap B) \subseteq A and A \subseteq (A \cap B)
So if we assume an element x \in (A\capB), then by definition, x\inA and x \inB. Thus we proved that (A\capB)\subseteqA.

In not quite sure how to prove the opposite, because if x is an element of A, that doenst necessarily mean that x is an element of A\capB...so i need help with the rest of it..or if you got any other ideas on how to approach it.

Thank you,
Emira!
 
Physics news on Phys.org
emira said:

Homework Statement


Prove that: A is a subset of B if and only if (A intersection B)=A


Homework Equations





The Attempt at a Solution

I tried proving the right side, that is

(A \cap B)=A
For two sets to be equal then they have to be subsets of each other...so:

(A \cap B) \subseteq A and A \subseteq (A \cap B)
So if we assume an element x \in (A\capB), then by definition, x\inA and x \inB. Thus we proved that (A\capB)\subseteqA.
Very good. That is exactly right!

In not quite sure how to prove the opposite, because if x is an element of A, that doenst necessarily mean that x is an element of A\capB...so i need help with the rest of it..or if you got any other ideas on how to approach it.

Thank you,
Emira![/QUOTE]
For the opposite, notice that you haven't used the hypothesis that A is a subset of B. If x is in A, then, because A is a subset of B it is also in B. Since it is in both A and B, it is in A\cap B
Now you have to prove the implication the other way: If A\cap B\subseteq A then A\subseteq B.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top