Speak now
Please Wait Image Converting Into Text...
Embark on a journey of knowledge! Take the quiz and earn valuable credits.
Challenge yourself and boost your learning! Start the quiz now to earn credits.
Unlock your potential! Begin the quiz, answer questions, and accumulate credits along the way.
General Tech Learning Aids/Tools 2 years ago
Posted on 16 Aug 2022, this text provides information on Learning Aids/Tools related to General Tech. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.
Turn Your Knowledge into Earnings.
I was once told that one must have a notion of the reals to take limits of functions. I don't see how this is true since it can be written for all functions from the rationals to the rationals, which I will denote ff, that
Since, as far as I know, functions like ∣x∣∣x∣ and relations like << can be defined on the rationals. Is it true you couldn't do calculus on just the rational numbers? At the moment I can't think of any rational functions that differentiate to real functions. If it's true that it isn't formally constructible on the rationals, what about the algebraic numbers?
Edit
Thanks for all the help, but I haven't seen anyone explicitly address whether or not we could construct integrals with only algebraic numbers. Thanks in advance to anyone who explains why or why not this is possible.
Derivatives don't really go wrong, it's antiderivatives. (EDIT: Actually, the more I think about it, this is just a symptom. The underlying cause is that continuity on the rationals is a much weaker notion than continuity on the reals.)
Consider the function f:Q→Qf:Q→Q given by
This function is continuous and differentiable everywhere in its domain. If x<πx<π, then there's a neighborhood of xx in which ff is a constant 00, and so it's continuous there, and f′(x)=0f′(x)=0. But if x>πx>π, there's a neighborhood of xx in which ff is a constant 11, so it's continuous there too, and f′(x)=0f′(x)=0 again.
So the antiderivatives of font-style: inherit; font-variant: inherit; font-weight: inherit; font-stretch: inherit; lin
No matter what stage you're at in your education or career, TuteeHub will help you reach the next level that you're aiming for. Simply,Choose a subject/topic and get started in self-paced practice sessions to improve your knowledge and scores.
General Tech 9 Answers
General Tech 7 Answers
General Tech 3 Answers
General Tech 2 Answers
Ready to take your education and career to the next level? Register today and join our growing community of learners and professionals.