Recent content by smithnya

Whoops I meant if the range is equal to the codomain then the function is surjective.

I understand the concept of a surjective or onto function (to a degree). I understand that if the range and domain of the function are the same then the function is onto. My professor gave an additional definition which I did not understand. Here it goes: \forally\inB \existsx\inA...

Homework Statement Find f°g for each pair of functions f and g.Homework Equations f(x) = {x+1, if x≤0 or 2x, if x>0 g(x) = {2x, if x≤-1 or -x, if x>-1 The Attempt at a Solution I am able to determine the following: (f°g)(x) = {2x+1, -2x, -x+1 However, these functions carry intervals, and...

I am having a difficult time wrapping my mind around the differences between a codomain and a range. Could someone explain the difference between the two and possibly provide an example?

But how can I just switch from ≥ to ≤? Isn't the crux of the problem that the relation be specifically ≥ on N?

So maybe that is where I am doing something wrong. For example: 1≥ 1 would make it reflexive 1≥2 and 2≥1 would make it not symmetric but 1≥2 2≥3 and 1≥3 is clearly not true, so why is it transitive is the statements are false?

Ok, is it allowed to substitute values for x and y , or a and b, etc to facilitate?

Ok, I am barely beginning to understand the subject. I understand that the relation ≥ on N(naturals) is reflexive, not symmetric, and transitive. I don't understand why it is transitive though. Can someone explain? Also, I understand why x2 = y2 is reflexive and symmetric, but I don't...

Hello everyone, I'm beginning a research project for a Math History class on Archimedes. Since technically it is a math class, I have to demonstrate and explain some of his math. Naturally, I can't explain and go into detail on everything the man did, and having only about a month to prepare...

You are right dot.hack. I meant > not =. Thanks for your help. Hodgey, the only think I can think of to bring the proof full circle is to reaffirm that (k +1)! > (k+1)2. What am I missing?

You mentioned in your previous post that we need to make our n = k+1 form resemble the original n!=n^2 form. Is that correct? Once we have k! > k+1, I really don't have an idea about how to do that.

Ok, yeah, initially I should have used n= 4, I believe. In the case of k!(k+1) = (k+1)(k+1) is it permitted to "factor" out an (k + 1) term, leaving us with k! > k+1? It would hold true for n≥ 4, right?

Homework Statement Prove that n! > n2 for every integer n ≥ 4, whereas n! > n3 for every integer n ≥ 6. Homework Equations See above. The Attempt at a Solution Ok, I am attempting an induction proof, but I seem to be stuck in the following step. In any case, I don't even know if...

Thank you so much. That was very simple and it explained what I needed to know. Nice avatar by the way. I am listening to "The Great Gig in the Sky" as I type.

Hello everyone, I have a simple question, but I am unsure. I know from a point p0 = (x1, y1) and a vector v = <a, b>, I can obtain a parametrized set of equations for a line in space such that x = x1 + at and y = y1 + bt. How can I check that any other point, not p0, is/isn't in that line?