INJECTIVE, SURJECTIVE, and BIJECTIVE FUNCTIONS - DISCRETE MATHEMATICS
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- Опубликовано: 6 сен 2024
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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)
I've got to say, you are far better than the professor I have for Discrete Math, it has everything to be quite the interesting, but some people just cannot teach. Anyway, thanks for the videos
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Yes!
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Discrete Math is so fun, when i can understand it
Thank you Sir, for helping me pass my exams
Injective: 0:39
Surjective: 5:25
Bijective: 10:07
Life saver
Life saver
İ'm trying to learn sth from the professor at the uni but it's impossible,because he cannot explain it clearly.You're doing a great job and now i'm starting to learn from your videos and it's really awesome.İ've never got functions before better than now.Thank you so much,maybe we can improve ourselves because of the person like you.
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i have to learn functions, uncountable vs countable sets, and relations in 2 days before my exam, wish me luck
f: R->Z given by f(x)=5x+2 is NOT surjective. It's clear just from looking at it that the range of f(x) is R, but the codomain of the function is Z, so it is not surjective. If, on the other hand, you're talking about the function f:R->Z given by f(y)=(y-2)/5, this again is not surjective by the same reasoning. For either of these functions to be surjective, you'd need f:Z->R
I was looking for this comment, Thank you
Got confused there at 9:29
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How'd it go?
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I watched this video and went back to my professors lecture notes. The first time I went through the notes, I felt like I only truly understood 60% of the material. After watching the video, I feel like I can now teach the course.
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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Only Passed?
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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 ?
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I have watched most of your Discrete Math tutorials. They are easy to understand and well arranged. I like your tone of voice which in not monotonous for a tutor. So Thank you and I hope you would develop more contents in the other areas. you greatly helped me in my last DM test.
"x2 was just x1 in disguise". This sentence alone clarified everything to me. My private tutor couldn't explain it to me in an hour. I assumed that if we were to put different numbers like 1 and 2, the outcome will be different. so f(1) not equal to f(2). Now I understand that f(x2) can not be different, and that is essentially what we had to prove. Thank you very much.
Been struggling to understand this for a few weeks now, this one video I accidentally found just explains it all. Thank you!!
For 10:06 How can it be R -> Z? Most x's will not be able to be mapped so it cant be a function.
I know you asked this question a long time ago, but here is the answer.
the function f:R->Z is a function. I will give an example:
f:R->Z, where x maps to the lowest integer except if x is an integer.
So this function will map any half open interval [x,x+1) to x itself. Meaning that the f(2.22245)=2 and f(10.999838)=10.
This function is clearly not injective, since we have a continium of numbers that maps to a single point. But the function would actually be surjective, since we maps to all of Z.
To give a better example, consider the function g:R->{1}. This function take any x in R, and maps it to the singleton {1}. We se agian that the function is not injective, but it is surjective. Functions defined like my to examples are still functions. An exercise for you to show would be:
Let f:R->R be a function defined as f(x)=1.
1) Prove the function is not injective nor surjective.
2) Now define the function f:R->R\A with f(x)=1. Choose A such that f will be surjective.
Hope this help.
@@nickgismo Hello. Why did you assume that the function does that? As far as I see from what is written that there are no special requirements for what you just described. And for the question you asked: part (2) Should the answer be R- Complement 1? Or how do you mathematically write that ? Thanks!
@@petruspetrusuom3342 Whar do you mean with I assumed the function did that? I really cannot see what you meant with that question. For the second part, you are correct. We would write it as: R\{1}^C or R-{1}^C where C denote the compliment.
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]
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
May God massively bless you sir.
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.
nice tutorial,but could you please give solution to inverse of f(x)=5x+2?
Man, you know i'm really reaching to enjoy math when I laugh at 1:12 when he says it's just x1 in disguise. Why was that funny... idk...
Everyone loves it when the killer is finally unmasked!
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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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You're welcome
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LING courses. Get your B-Socs 😂🙃
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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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You made this sooooo much simpler to understand than my University Professor who gets paid $100K+ yet still can't explain a simple concept simply. Kudos to you sir 🫡
People are talking about uni while I’m learning this in secondary school
This material should go on Khan Academy since they don't have a discrete math course and these are excellent
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i swear my professor skipped three whole sections. We had a test on sets last week and now we have one due next week on this thing and this is like your 54th video in the playlist
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Please from your illustrations on injective functions how come we got the + or -X1=+ or -X2?
Thanks a lot!!! Wish I had found you earlier
amazing! much better than my professor who explained this for 2 hours but still confused me
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I gues 09:36 is not a function in the first place since there are some x's that cannot be mappen on to a value in our codomain. Say x=1.231 which is in our domain. However, when x takes that value, y happens to be a value that is not an integer, and I suppose that means what we have is not a function.
you are an absolute god! You are better than my lecturer man!
Have you ever dealt with quasi-injective functions? This is a function f:A -> B where f(x)=b has at most finitely many solutions. Does it have any applications? Yes, it has. Consider a convergent real sequence a_n that converges against 0. Then a_n = a can have at most finitely many solutions if a /=0.
So my understanding is that injective basically means that every f(x) has its own unique y, but not every y can be reached by f(some x), whereas surjective means that every f(x) has its own unique y AND every y can be reached by some unique f(x).
But I don't understand how bijective is different from surjective. If every f(x) has its own unique, y, then it is injective. In a surjective function, every f(x) has a unique y, therefore every surjective function is also injective. In a bijective function, a the function must be both surjective and injective, therefore, because every surjective function is injective, every surjective function is also bijective. What is the point of even having bijective if it is the same as surjective? What am I missing?
Your videos are great and easy to understand but if you post some exercises on the discription it will be more helpful. 👍
For the function f(x)=5x+2, is giving two examples really proving it? If asked on a test, I have a feeling that won't be sufficient.
Thank you, my learning material did not explain at all how to determine if a function is injective, surjective, or bijective.
Glad it was helpful!
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?
Is the process for proving injective and surjective the same for piecewise defined functions? Only you have to do a case for each piece?
i can't understand a word that comes out my professors mouth but random youtube guys make perfect sense lol
The best explanation ever.
Great work sir! However, I think you misused R as Rational when it is Real Number . Q is denoted as Rational.
I think he assumed we knew Q is a subset of R
This video is great, however i just relazied something and would like to let everyone else know about it incase you are having a hard time understanding on how you can decide if something is surjective or not. You follow his steps, he explains perfectly on how you can determine if it's surjective or not. But you must pay attention to the range/domain, the domain/range decides whether the function is surjective or not. If the range is Z -> Z, then you can create a function (for example f(x)=x-4, the inverse is y+4 = x) which will give an output that ranges from all negative and positive values. If the inverse functions gives an output of a value that does not belong in the range/domain then it is not surjective. For instance if the function output all real numbers, or fractions then it is not surjective. I hope this helped, as this helped my stupid ass to understand lol.
Man I can't wait to be done with this discrete math stuff.
Ur really good keep up the good work my g
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!
Thank you for the worksheets and solutions.
Thank you so much❤
The best video ever !! I am now crystal clear. Thanks.
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Can bijective function have unused elements in domain?
I mean, does there exist such x in X such that f(x) >< y?
I mean it's possible, that set Y has less elements than set X.
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????
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So to check if a function is invisible, I have to check if the function is injective and surjective (bijective)?