INJECTIVE, SURJECTIVE, and BIJECTIVE FUNCTIONS - DISCRETE MATHEMATICS

[@SgtMajorJay]

Seriously... that's it?? Why was this to complicated when our professor taught it. Thank you so much!

[@jackmenirons4989]

My thoughts exactly. After struggling through bad lectures and cryptic textbooks these 5-10 minute videos make it as easy as addition and subtraction. No idea why this seems to be so hard to teach in most colleges.

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Hahahhaha could relate!

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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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[@TheMrsteakman]

Injective: 0:39 Surjective: 5:25 Bijective: 10:07

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Life saver

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Life saver

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Saved their life 2x

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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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[@Trevtutor]

You're welcome

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LING courses. Get your B-Socs 😂🙃

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"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.

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Been struggling to understand this for a few weeks now, this one video I accidentally found just explains it all. Thank you!!

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Thanks!! This is really way more understandable and intuitive than what my professor had taught in class🤩

[@RyanTorrecampo]

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.

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[@jamesbaldwin7169]

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

[@realitymeetstheworld8217]

I was looking for this comment, Thank you Got confused there at 9:29

[@umarfarook-ng7rs]

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[@danielnwosu3971]

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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I have an exam tomorrow, and this is really helping me. Thank you!!

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Why do they have to make everything so difficult. Thanks discrete maths for making my life easier.

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[@adamboyen4727]

Very helpful on my assignment to state that an equation is neither injective nor surjective, thank you

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[@brandonwootton5575]

9:40 wouldn't an irrational input make this not surjective

[@FyroozAhsan]

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[@damianpatel2794]

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?

[@Bikash_The_1]

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.

[@emreturker4175]

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.

[@gospelmusic24]

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

[@supraocny]

Thank you, my learning material did not explain at all how to determine if a function is injective, surjective, or bijective.

[@Trevtutor]

Glad it was helpful!

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[@DarkCIoudStudios]

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.

[@nickgismo]

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.

[@petruspetrusuom3342]

@@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!

[@nickgismo]

@@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.

[@riffsfromtheabyss666]

15:50 Reminds me of double negation

[@subhayansinha4417]

This was helpful. Thank you sir.

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Thank you for all the videos, they're awesome!

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Thanks for the video sir. I found it's helpful...!!

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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...

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Everyone loves it when the killer is finally unmasked!

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[@sahanaav-ex5wb]

iff means if and only if

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[@fairlymoon448]

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]

[@abhi-nav22]

Your videos are great and easy to understand but if you post some exercises on the discription it will be more helpful. 👍

[@Chrisymcmb]

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 ?

[@patrickmensah2338]

very understanding video. Thank you so much!

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[@yasserzahran9921]

i shall say thank u for the explanation

[@nixonangatiaamukoya2616]

nice tutorial,but could you please give solution to inverse of f(x)=5x+2?

[@jessicajaja6132]

I was honestly failing Discrete maths until I watched this video. Thank you

[@BoonkiCoC14]

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.

[@ngndnd]

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