Injective Functions (and a Proof!) | Injections, One to One Functions, Injective Proofs 

My asnwer: I would say no because if x is a negative number then it maps to the same number of its positive counterpart, thus the function does not preserve distinctness

Absolutely love the videos! keep the great work up. you deserve much more subs! would love to see more proves on maybe a little harder surjective, injective and bijective functions

i got it when you plug in numbers instead of putting letters like x it makes me confusing thank you it's really helpful because there's so many tutors in youtube but i find it confusing until one time i watch your videos it makes me a lot easier

Hi, your videos are very helpful. Are all videos organized in playlists? In which playlist can I find this for example? I wanted to watch the whole series in which the concepts bijective, surjective, injective are explained. I counted the number of videos and the number of videos in the seperate playlists, and it looks as if there are more videos seperately than videos, organised in playlists? Can you put a link in the description below the videos or in the title, or anyhow make it possible to choose the whole playlist, once we find a video, we are interested in. Thank you very much!

3:04 Also, if one of the elements in A mapped to something outside of B it also wouldn't be injective anymore right?

Thanks for this lesson, I stopped at this section before I went to bed and was like OMG jeeeze, but this made reading that section a lot easier, It was just like symbols and new words everywhere! haha

5:25 It isn't injective because the function maps to the same number once, making it not longer distinct, or one-to-one?

F(x)=3x^2 F(-x)=3(-x)^2 F(-x)=3x^2 since F(-x)=F(x) 3x^2=3x^2 Then f is injective 3x^2=3x^2

but sir it seems the proposition dimension is not independently efficient to prove injection Both proposition and contrapositive should be used concurrently

Hello can i ask how did g(-3) became g(3) in 5:19

if x!=0; x != -x ...(a) ; However, 3(x)^2 =3(-x)^2 -->f(x)=f(-x) ....(b); based on (a) and (b) --> f(x) is not injective

Nice work.. thanks...

Sir it is not injective as -3 isn't equal to +3 but their images are... Hence defying the contrapositive

f(x) = 3 x^2 is not injective since f(-1) = 3. (-1) (-1) = 3 = 3. (1). (1) = f(1).

0:55

i got the answer right

Hi