Recent content by 1800bigk

yes, try to find something that you enjoy doing and can get paid for it. It's not a good idea to become an xyz just because an xyz makes so much money. If you dread getting up and going to work everyday then you will be miserable no matter what the salary is.

doesnt [sqrt(a)]^2 = |a| ? maybe that's just in the reals

yes that's right. chain rule. no matter what positive integer n, in the end you will have some fixed number in the numerator but the denominator goes to infinity so it wins out and the limit will be zero.

i think I am right, I am asking because some of the kids in my class said zero is not an integer and they said i should of picked two distinct elements to show its not closed but i said it didnt matter.

Let S = {R-Z}, the set of all reals that are not integers. Is S a subring of R? I think not because 1/2 is in S but 1/2-1/2=0 so S is not closed under subtraction so is not a subring. is that right?

First off I would say not to take a semester off. Secondly, you have to ask yourself how much do you like math and how much do you like physics? Have you thought about engineering, that is a good mix of physics and math and its only one major. If you really love math and don't mind spending...

a removable discontinuity is just something that we can fix or adjust to get the function continuous. It usually means a function is discontinuous at some point or hole in the graph and all we have to do is plug the hole if you will, or redefine the function at the point in question. The...

square both side and see if you notice anything and remember if A=A then, surely A is less than or equal to A

what are the points of the graph you want to find an equation for?

what kind of requirements do I need? Do I need to say something about Xn converging to an element in S? Do I need to use subsequences?

I was just getting ready to say just because S is bounded doesn't mean S is closed and I need S to be closed to go the easy way. I supposed f is uniformly continuous and unbounded then I said since S is bounded then there is a convergent sequence {Xn} in S by bolzano and since {Xn} converges...

Prove that if f is uniformly continuous on a bounded set S then f is bounded on S. Our book says uniform continuity on an interval implies regular continuity on the interval, and in the previous chapter we proved that if a function is continuous on some closed interval then it is bounded...

the piecewise functions above are both zero divisors. since neither function is the zero function but their product is zero for all x, ok I think I get now. so any function that is not the zero function , but takes on the value of zero for some x, is a zero divisor. Becasue if my function f...

so are there any zero divisors

The problem says under normal function multiplication. The zero function is Z(x)= 0 for all x