# anon

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bio website location Las Vegas, NV age member for 2 years, 7 months seen 3 hours ago profile views 63

 12h comment What is wrong with my contradiction? @DanielFischer: Not really understanding the transfinite stuff, but I'm sure I'll get to it someday, otherwise your argument makes sense. I'll also be sure to look into this "Lebesgue integral" someday too, looks interesting. 12h comment What is wrong with my contradiction? @MichaelJoyce: Ahh, ok, I helps me to remove the notion of an infinity element from the natural numbers and just think of it terms of finite, that then it makes more sense for induction to be forced to work for only finite cases. I still throw infinity out there way too much, I need to start avoiding it more often. Haha. Thanks! 13h comment What is wrong with my contradiction? @MichaelJoyce: Thanks, I guess the real problem is I am just looking for a more obvious reason why my inductive step doesn't work. Because if I didn't know about the proof that Spivak gave, I would never of saw anything wrong with the idea of doing this. I think I might need to just look into induction more to have it really soak in. :) 13h comment What is wrong with my contradiction? @DanielFischer: It's interesting now because since I never realized that, it's kind hard to think why it works with only a finite number of integrable functions. It would be nice if you could maybe elaborate more as to why this is true, although, maybe I need to just stare at the definition of induction longer for it to make sense. 13h comment What is wrong with my contradiction? @Daniel Fischer: Hmmm, didn't know that induction only works on finite sets... That would cause this to fail if that isn't true... Thought the whole point of induction was to be able to do stuff with infinite. Well, that was rather a silly mistake on my part. Apr11 comment Preventing “proof by homework”? @zyx: Proof by homework is when you say: "Theorem X is true because the question says it is." When I put the step about multiplying (a^n + ... b^n) * (a + b) I made it equal to the correct thing. Sure my "proof" is correct, but it was not thorough and I was left unconvinced that multiplying (a^n + ... b^n) by (a + b) actually resulted in a correct expression. Apr11 comment Preventing “proof by homework”? Thanks! I wish I could accept both yours and mayrb's answer. :( Apr11 comment Preventing “proof by homework”? Ahhh, thanks for reminding me of that fact, yes I did prove it in previous answers. I ended up using a combination of both yours and David's answers. Apr8 comment Getting different answer when evaluating an integral from a released exam. Multiply by a ... not 1/a Apr8 comment Getting different answer when evaluating an integral from a released exam. Actually, I forgot to add C to the second integration... darn it. XD Apr8 comment Getting different answer when evaluating an integral from a released exam. I'm glad it wasn't me forgetting a rule. I get it. Multiply by 1/a and k = aC1 - aC2. Thanks! Hahaha. Apr7 comment Help evaluating summation. @Ron Gordon: Thanks! If you post a confirmation (as an answer) to my question I'll be happy to accept it! Apr7 comment Help evaluating summation. Yeah. It is 10^(i/n) Nov22 comment How to prove the divisors of two numbers is the same as the divisors of a and b? Ok, so I think I get it: $\gcd(a,b)$ is the smallest linear combination solution. All divisors of $a$ and $b$ divide the linear combination. Therefore, the two sets of divisors are the same then? Dec24 comment Is a ring with the following properties semiprime? (Part 2) You should try merging your accounts. Dec6 comment Row-sum for an integer triangle w/o brute force @AndréNicolas: Ok, thanks a lot! Dec6 comment Row-sum for an integer triangle w/o brute force @AndréNicolas: I don't get the second equation: $\frac{(n-1)(n)}{2}$... What is that doing exactly? I know the first one is summing up consecutive integers but I don't understand exactly what the slightly modified one is doing. What exactly is it doing? Oct25 comment Need help finding: $\frac{d}{dx}\frac{\sec{x}}{1+\tan{x}}$ @ArturoMagidin: :0 Omg. I must be retarded. Consider my mind blown. Oct24 comment Need help finding: $\frac{d}{dx}\frac{\sec{x}}{1+\tan{x}}$ @Aturo: I took your advice and didn't do that, but I still got the wrong answer. Sorry if I made another obvious mistake. Do you know what is wrong with my logic this time? Oct24 comment Need help finding: $\frac{d}{dx}\frac{\sec{x}}{1+\tan{x}}$ @Aturo: O wait, I see what you are getting at... Ok, I'll try again and see how it goes.