Recent content by Mr. Fest

Great! It's because B is the range of A, and the function f: A --> B is bijective (all elements of B is "used"), therefore the function h: B --> C, where B is the domain will also be bijective. This is what springs to mind at first thought. But I'm not completely certain. What I meant...

Well, f(x) = x is such a function. Because then the image of A is always on A for all x \in A. And we have shown reflexitivity... Also, it is both injective and surjective as we have f(x1) = f(x2) --> x1 = x2 and since every element in range is the image of one and only one element in the...

You asked me to find a bijection so I did. Why not work it out for me? Might understand better if you do...if it's not too much to ask.

I'm at work now and I'm not sure if the solutions chapter states any thing like that. But for example f(x) = 2x; x \in R then f: A --> B is bijective. We have a one-to-one correspondence. It is symmetric since f: A --> B; f(x) = 2x. f: B --> A; f(y) = y/2. f: A --> B; f(x) = 2x. f: B --> C...

Since the function A --> B is bijective you have only one image on B for every element in A and also the function A --> B covers all elements of B. So, if the same is given for B --> C then per definition A --> C and the function is transitive. I'm assuming this is correct? And if the...

But if two sets are equal, aren't they also equal in size...

A = {1, 2, 3, 5, 7, 9}; B = {1, 2, 3, 5, 7, 9} then A is perfect subset of B correct? And if it is a perfect subset then the set A is equal to the set B, correct?

What I meant was that if f: A --> B and A is a perfect subset of B then A = B and ergo A goes to A. Example, y = x + 1 A = (-inf, inf) and B = (-inf, inf) then A = A and A --> B can be seen as A --> A. Is this what you mean/meant?

Isn't surjection that A = B and therefore the range of f is equal to B and domain of f-1 is also equal to B. Just to make it clear, "same powerfulness" has not yet been mentioned in the textbook. I'm doing a chapter on "Functions and relations" where the focus has been on injection...

But if it is a bijection A = B ergo A goes to A? Or am I thinking the wrong way?

Homework Statement We say that two sets A and B have the "same powerfulness" if there is a bijection from A to B. Show that the relation "have the same powerfulness" is an equivalence relation between sets. Homework Equations An equivalence relation satisfy the following: xRx...

Sorry Ray, totally forgot about that suggestion. Will look it up though! Thanks!

Yes, thank you. I believe I mentioned I know that I can prove it in the same manner I can prove sqrt(2) is irrational. I just wanted to know if I can prove it using the fact that sqrt(6) = sqrt(2)*sqrt(3) and the fact that sqrt(2) is irrational...

Haha my bad...but still then p^2 - 1 = 24... But 8 = 2*4...so if it is divisible by 2 and 4 it is divisible by 8 and also divisible by 3 --> divisible by 24? But still. We have shown that it is divisible by 8 and 3 why does it become divisible by 24? Sorry if I'm slow and tired at the moment...

Remember that p is a prime number > 3 so the smallest p > 3 is 7...