this definitely helped me wrap my head around proving surjection before my discrete structures test tomorrow! as did your injection tutorial. thanks a bunch.
@jamesr2784
8 жыл бұрын
WOW. Very thorough and well explained. Thank you sir.
@acdude5266
2 жыл бұрын
Great work
@vincentpi67
11 жыл бұрын
Enjoy your videos a lot, hope you will continue to post.
@shreyaroy3179
6 жыл бұрын
Thank you for such an easy explanation!😊
@maed1917
10 жыл бұрын
this was so incredibly helpful! thank you so much!!
@ilona7051
6 жыл бұрын
This is great! Really helpful, thanks!
@GVSUmath
11 жыл бұрын
5/2 doesn't equal 5, but f(5/2) does equal 5. The function f operates by stripping off the numerator and returning it as the output. Under this function f(5/3) would also equal 5, and so on.
@58TaskForce
9 жыл бұрын
Thank you!
@WAAGOM
10 жыл бұрын
This is so helpful. Out of curiosity, what book do you use for your class? I might recommend it to my professor. We use "The Art of Proof", and I don't find it very useful.
@GVSUmath
11 жыл бұрын
Watch the video more carefully. It doesn't claim a/b = a. It says f(a/b) = a. There's a function involved.
@MrJewar4
4 жыл бұрын
Can the relation defined as f(a/b)=a be called a function? What I'm trying to say is that if you assume f is a function, you can call a/b=1/2=2/4. Then f(1/2)=1 and f(2/4)=2, but 1/2 = 2/4 and f(1/2)≠f(2/4) !
@ra7vn809
8 жыл бұрын
Thanks for the vid but at 10:50 why would you claim a natural number to be equal or greater than 1? Why ignoring 0?
@varunmanavazhi5664
8 жыл бұрын
+Ra7vn 0 is not a natural number. Natural numbers are 1,2,3,4.. and so on.
@guthriekaruma1356
6 жыл бұрын
natural numbers are counting numbers. We start counting from 1 not zero
@Seastric
5 жыл бұрын
For some novices, you could have done them the world of good if could have differentiated the meanings of the co-domain range in your definition, the majority tend to think they are one and the same.
@jamesplank9548
10 жыл бұрын
This is rather abstract in thought so it is understandable that it would be hard to believe that f(5/2) can indeed equal 5. But indeed functions can do crazy things :)
@TheGamerGameplay1
6 жыл бұрын
James Plank Isn't so hard. You have 5/2, and you have something called f. When you use f in some number, the number is multiplicated by 2. So, using f in 2, you get 4, because thats what "f" do. Then, f(2) = 5, because f applied on 2, returns the double of the number inside the parentheses. Therefore, f(5/2) will return the double of 5/2. So, f(5/2) = 2*(5/2) = 5
@TheGamerGameplay1
6 жыл бұрын
I hope you get it. If not, just reply. Good luck
@ashleylove6840
3 жыл бұрын
@@TheGamerGameplay1 I Still dont get it :(
@aurangajebazad4751
4 жыл бұрын
Onto means every range will be associated with domain that is comdomain associated with domain,but in this case u just proved that one domain is associated with a range(subset of codomain) so how can we say that we proved it onto!?
@jasonwilliford9877
10 жыл бұрын
Nice video. One comment: your first example of f is not well defined. You should specify some form that rational numbers must be written in before computing f, like "lowest terms with a positive denominator". Without that, f(2/4) = 2, but f(1/2) =1, for example.
@GVSUmath
10 жыл бұрын
Thanks Jason. In the context of the book we are using for the class, the rational numbers are defined specifically so that the numerator and denominator are relatively prime. So students would know that this is well defined -- however it's a fair point because you guys don't have that context. Duly noted.
@TheBoondokSaint
9 жыл бұрын
GVSUmath I am a complete novice at math, and before I share my thoughts on the proof, I do want to say thank you for publishing the video. Thinking in terms off 'targetting from the codomain' is quite helpful. My thought is: I don't think a definition of rational numbers as relatively prime numerators and denominators is sufficient. For example, if m= 3/2 and n = 3/5, both numbers are integer numerators over integer denominators in simplest form with denominator greater than zero where f(a/b) = a, but your proof in this video only covers m. Would you not need to state the definition of a rational number and reframe the denominator as a variable integer greater than zero and relatively prime to the numerator?
@pajohns
11 жыл бұрын
I'm not sure I understand this - at 4:46 you mention that n/2 maps to n. In your example you use n = 5. 5/2 does not equal 5. Wouldn't it be 2n/2 maps to n?
@alexs2415
10 жыл бұрын
How does 5/2 give you 5?
@davidsteinberg4909
10 жыл бұрын
5/2 gives you 5 if you apply the function f to it. Remember: f(a/b) = a, so f(5/2)=5
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