INJECTIVE, SURJECTIVE, and BIJECTIVE FUNCTIONS - DISCRETE MATHEMATICS
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- Опубліковано 18 тра 2015
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We introduce the concept of injective functions, surjective functions, bijective functions, and inverse functions.
#DiscreteMath #Mathematics #Functions
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![[Discrete Mathematics] Functions Examples](http://i.ytimg.com/vi/Kn-LrdgHVhQ/mqdefault.jpg)
![[Discrete Mathematics] Functions Examples](/img/tr.png)
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Injective: 0:39
Surjective: 5:25
Bijective: 10:07
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My professor has tons of practice questions and explanations and I still couldn't understand how he was doing the injective proofs. I wish he had just prefaced from the start, like you did, that he is using the contrapositive of the definition of injection. I have my finals tomorrow in an introductory discrete math course and you might have just saved my grades, ty.
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Very helpful on my assignment to state that an equation is neither injective nor surjective, thank you
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Is the process for proving injective and surjective the same for piecewise defined functions? Only you have to do a case for each piece?
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Why is y on the right and not the left in F: R->Z? You originally wrote y=x so I assumed that is the order. I'm going to have an exam so I would like to know if you just wrote it that way or if that's standard. Thank you so much!
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Sir I have one question please make video on it
as you have said we assumed x1=x2 then f(x1)=f(x2),,then in what case it will not satisfy the condition...I meant to say by this method how can we show a function is not injective
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THANK YOU VERY MUCH SIR, FOR TAKING YOUR PRECIOUS TIME TO PRESENT THIS TUTORIAL. PLEASE SIR, I AM NEW TO THIS KIND OF MATHS, HERE ARE SOME QUESTIONS, I NEED YOUR HELP ON THIS ONE...
1. (a) The universal set is {1, 2, . . . 10}. Let A = {1, 2, 3, 4, 6, 7}, B = {2, 3, 4, 5, 8} and
C = {1, 3, 5, 6, 8, 9}. Find the elements of the following sets:
(i) A∆B
(ii) (A ∩ B)
′\C
(b) Given the relation R such that R = {(m, n) ∈ R|m, n ∈ A, m2 − n ≥ 4} when A
is the set {0, 1, 2, 3, 4},
(i) express the relation R as a set of ordered pairs.
(ii) draw an arrow diagram to represent this relation.
(iii) investigate whether the relation is symmetric, transitive and/or reflexive.
(c) Determine if the following functions are one-to-one and/or onto and use these to
determine whether the function is a bijection. Assume both functions are such
where f : R → R.
(i) f(x) = 6x − 9
(ii) f(x) = x
2 − 2x + 1
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Okay, So if instead you were going from Z->R, in your example y-2/5= x how do you do that? I understand when its Z->Z or R->R, and R->Z makes sense because theres only a limited number of integers. but Z is everything except C numbers. so is it considered still onto? how would you do the math in the equation to prove it? [your vids rock even if i dont get everything]
My doubt is still related to the size of the domain. If I have a finite set A with unknown values, can I say it has a bijective relation to N because it is countable, or I can say it is countable only because it is surjective, because saying it is bijective would mean the size of N is the same as the size of A????
This material should go on Khan Academy since they don't have a discrete math course and these are excellent
At around 9:45, you say that with a domain of real numbers and a codomain of integers, it is surjective; however, some domain values will not even have a mapping/value. Doesn't this mean that it no longer qualifies as a function?
You can simply say co domain is subset of range.Or range(R)is bigger set than of co domain (Z).Hence That's not even a function.
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nice tutorial,but could you please give solution to inverse of f(x)=5x+2?
Thank yu Sir
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So is it true that all surjective functions have a domain and co domain with the same cardinality? Or not because a function could have a one to many mapping?
All other discrete math videdos are trash.. these make sense!! Thank you!!
For the last example of the surjective proof for f(x)=5x+2, we wrote the function as y=5x+2, then rewrite the function in terms of x, so x=(y-2)/5. We lastly tried to prove if we mapped f:real->integers. This does work if we plug in for the function in terms of x, but if we plug in this mapping for the function in terms of y, then it is not possible. Should the mapping not work for the function in terms of a and y both when mapped f:real->integers ?