I believe you have it backwards sir, but there is no harm in this confusion. Definition of injective function: A function f: A →B is said to be injective if for all x,y in A, if x ≠ y then f(x) ≠ f(y). Contrapositive version of definition: A function f: A →B is said to be injective if for all x,y in A, if f(x) = f(y) then x = y. In practice we ( us mathematicians) tend to use the contrapositive version of the definition in proofs, because we can 'do algebra' more easily with the = sign (harder to do algebra on ≠ ).
when is doing his first example to prove that e^x is injective, how can we apply a natural log to both sides? Doesn't that assume that the natural log function is itself injective? Is this circular reasoning?
+David Schorr I agree. Instead, he should have multiplied both sides by exp(-x2) to obtain: exp(x1-x2)=1. We can do this since we know unique inverses exist in non-zero Real numbers under multiplication. Then, we conclude this is true if and only if x1-x2 = 0. Therefore, x1=x2.
Isn’t the point just to use algebraic manipulation to find x1 = x2 ? Why do we need to worry about such small things lol... because in multiplying both sides by another e^-x2 tbh that seems far more circular to me. Since you’re still using the same problem they initially gave you to attack the problem at hand. If both techniques are just algebraic manipulation being used to force a specific conclusion to be displayed, then I honestly don’t see the difference.
please what software are you using for the teaching, the screen recording and the on screen writing, please I like it and wouldn't mind if you could introduce it to me
Given a function f: A->B. Here is a list of useful definitions of 1-1 and onto: Un-equal inputs have un-equal outputs. 4:10 (Contrapositive): Equal outputs have equal inputs Another way of thinking of 1-1: Each element in the codomain is mapped by at most one element in the domain. Symbolically we can write the latter as, for all b in B, | f^-1(b) | = 1 Also: If for each element b in B there exists an element a in A such that f(a) = b , then f is onto. If f(A) = B , then f is onto. If the image of the domain of f = codomain of f, then f is onto.
I like your voice and your teaching style. That being said , 9:00 I don't like this proof. The definition of 'a function is not injective' is the negation of 'a function is injective'. By negating the universal quantifier, the negation is an existential statement. So, you need to produce a counterexample, which you did not. But you did produce a schema to choose specific counterexamples, like a template. And clearly | 2 | = | -2|, but 2 is not equal to -2 . That is a counterexample. You are not the only student/teacher to do this. a lot of students/teachers do this. And we need to stop it, because first order logic looks loosey goosey.
