Recent content by BrianMath

Homework Statement Verify that the inverse of an automorphism is an automorphism. Homework Equations The Attempt at a Solution Let f:G\to G be an automorphism. Then, f(xy)=f(x)f(y) \forall x,y\in G. Then, we define the inverse f^{-1}:G\to G by f^{-1}(f(x)) = f(f^{-1}(x)) = x...

Homework Statement Theorem 1.4: Show that every Cauchy sequence is bounded. Homework Equations Theorem 1.2: If a_n is a convergent sequence, then a_n is bounded. Theorem 1.3: a_n is a Cauchy sequence \iff a_n is a convergent sequence. The Attempt at a Solution By Theorem 1.3, a...

This would only prove one implication. This is an if and only if proof, so you have to prove that a=b \implies \frac{na + b}{n+1}=\sqrt[n+1]{a^nb} AND \frac{na + b}{n+1} = \sqrt[n+1]{a^nb} \implies a=b

No problem, little mistakes like these happen to the best of us. :smile:

Where did the x-1 in the denominator go? \lim_{x\to 1} \frac{\frac{1}{1+x^2} - \frac{1}{2}}{x-1} = \lim_{x\to 1} \frac{\frac{2}{2(1+x^2)} - \frac{1+x^2}{2(1+x^2)}}{x-1} = \lim_{x\to 1} \frac{\frac{2 - 1 - x^2}{2(1+x^2)}}{x-1} = \lim_{x\to 1} \frac{\frac{-x^2 + 1}{2(1+x^2)}}{x-1}

Okay, if I've read this correctly, so far you have \lim_{x\to 1} \frac{\frac{-x^2 + 1}{2(x^2 + 1)}}{x-1} right? What happens if you factor out a -1 from the top fraction's numerator? Is there any algebra you can do on that to divide out the x-1?

I think this could be made elementary by using \sqrt{2} in place of \pi, since the irrationality of \sqrt{2} is quite easy to prove by contradiction. :biggrin: But then I suppose using \pi for this proof also works to prove that there are transcendental numbers (a stronger attribute than...

Okay, here's my attempt. Hopefully it's correct. Tell me if it's similar to what you have. :smile:

Would you mind if I post my attempt at a proof? :smile:

Homework Statement If A, B, and C are subsets of the set S, show that A^C \cup B^C = \left(A \cap B\right)^C Homework Equations A^C = \{x \in S: x \not \in A\} A\cup B = \{x \in S:\; x \in A\; or\; x\in B\} A\cap B = \{x \in S:\; x \in A\; and\; x\in B\} The Attempt at a Solution...

Yeah, what reason should a parent, of all people, ever see to their child's well being? That's just absurd! Yeah! Let's just throw them under the bus and be done with it. All they ever did was cry when they were an infant, show their uninteresting macaroni art from kindergarten, and ask for...

The reason this doesn't work is that (1 - 1) + (1 - 1) + (1 - 1) + ..., which is a type of series called an Alternating Series, does not converge (meaning it doesn't add up to a finite number) and diverging series do not conform to the normal rules of arithmetic. An alternating series is a...

I think this only works when you're making a statement about all non-primes having those properties. That is, your counterexample must be against a statement that says "for all non-prime p, (a,p)=1 or p divides a". However, in your statement, you're only assuming that there might exist one such...

I think you mean (a+ 8y)x^2+ (6-6y)x - (8+ ay)= 0 on the last line. :smile:

I know this thread is over half a year old, but I happened to come across it while searching some stuff about Complex Analysis on Google. It's possible that you may have already gotten an answer somewhere else, but for others who would like to know why this is and may end up here after a search...