watch the next few videos! It gives us a way to represent logical expressions algebraically so that we can manipulate them, combine them, negate them, find simpler expressions with the same truth values, ...
Wow, those last 4 examples really cleared things up for me! Thanks! I was really confused on figuring out whether something is surjective, the simple statement: "do ∀ x∈X go to ∀ y∈Y" made it way easier!
Hello. Good video, one question though. Can a paradox be a WFF? I mean, the last example you put in 6:02 is a paradox in the sense that there is no q that satisfies that formula (for (p∧q) to be true, both p and q have to be true, and thus cannot imply that ﹁q is true)
"right inv iff injective" isn't true. Left inv iff injective does not require choice for the proof - you can pick a single element from a non-empty set without AoC if that's what you're worried about. Surjective implies right invertible is the one that needs choice - for each y in Y you must choose g(y) from the set of things f maps to y, and in general the AoC is needed for that. Right invertible implies surjective does not need AoC.
Depends on your tastes. A.O. Morris - Linear Algebra, Kreyszig - Advanced Engineering Mathematics, Halmos - Finite Dimensional Vector Spaces, one of the Gilbert Strang books, ...
Hi. I have just seen the video and I thought that, regarding the Euler's conjecture counterexample, we could look into all integers to see if there exists one positive and one negative number that satisy the condition to be a "blurred number" in the example you provided. To check this, we want the differece of both fifth powers to be 4196468331 (144^5 - 133^5 - 110^5) so with a while loop I realised that the numbers couldn't be (in absolute value) grater than 170 because 171^5 - 170^5 > 4196468331. Therefore, I just had to make 2 loops running from -170 to 170, or even better, one (e.g. the variable i) from 1 to 170 and other from -i+1 to i. This code gave me just the solutions already-known 84 and 27, so those may be the only integer solutions. This makes my doubt about if my check was actually necessary...
that's interesting! You can eliminate negative numbers from the problem by looking instead for solutions of a^5 + b^5 + c^5 = d^5 + e^5 - a good search term if you're interested in this stuff is "sums of like powers," it is a huge field of study in number theory. There are nontrivial examples of a sum of 5th powers equalling a 5th power where one of the 5th powers is negative, one of them is ✷^5 + ✷^5 + 6237^5 + 14068^5 = 14132^5
@@MarkySharky algebra 3 is next year, for you, so things may be different - though i doubt they expect you to remember a specific numbering of the vector space axioms
