Algebra: How to prove functions are injective, surjective and bijective

Math1141. Tutorial 1, Question 3. Examples on how to prove functions are injective.

[@sanjaym5284]

this really helped me a lot for my exam which is tmr , loved your explanation

[@crimson4066]

20:53 where did you get the "divided by a-5" from? x (5 - a) = -1 - 2 a, so x = (-1 - 2a) / (5-a)

[@hasibalirishad6940]

love you boss

[@Acesif]

Thank you so much, this tutorial is the most helpful of all the other videos I came across

[@promathacademy8512]

That's just crazy....but thank you very much for letting us know, please consider subscribing.

[@marcogimenez18]

You're the best! A REAL EXAMPLE

[@promathacademy8512]

We love and appreciate your support!

[@kimun2106]

Thanks this by far the clearest video , def subscribing

[@aayishaabusalih6043]

Thank you got a better understanding of this concept 👍

[@datawkwardweeb8303]

This was hella useful lmao Have a test next week Thank you so much

[@promathacademy8512]

Thank you for your comment. It was our pleasure.

[@IKdeoSSa_7]

Good video👍

[@dorajn]

exellent video, very transparent

[@mariiooP]

Great examples and excellent explanation. Thanks!

[@promathacademy8512]

Glad you found it helpful. Please consider subscribing to help the channel grow. Every subscription counts and we appreciate it so much!

[@amgadelgamal4445]

Amazing video!!

[@dimitrismarkopoulos9846]

Very useful

[@JPL454]

Proving that a linear transformation is bijective seens so much easier than proving that a normal function like y=x+3 is bijective for some reason

[@janahabachy9109]

i do not understand why or how did you divide by a-5 at 20:50

[@jubytr497]

This is what I was looking for.Thnx .Very well explained.

[@promathacademy8512]

Thank you for you immeasurable support.

[@radhwanalbedany8595]

for the question 3 / the function is surjective since its already written that R - {2} this is read as " for all R numbers except number 2"

[@danielraphael919]

Yes, that is correct!

[@promathacademy8512]

Love to see our subscribers interacting. Just to answer your question, your question is already answered ! Keep up the great work.

[@coast2coastpod806]

I think by modulus you meant absolute value, but the overall content is understood, thank you.

[@promathacademy8512]

That is correct.

[@thefxbro433]

in the last proof example why did u divide by a-5 instead of 5-a?

[@danielraphael919]

I asked the same question. He said it was an error.

[@jazzysocksdude]

How would this work for functions with multiple expressions, such as f(x) = { x + 1 if x is even, x - 3 if x is odd } ?

[@promathacademy8512]

This is a very good question. It is not injective - test y =1 with the horizontal line test. To prove surjectivity we have to know the co-domain. Send a picture of the question to [e-mail]

[@jazzysocksdude]

@@promathacademy8512 thanks, I've just emailed it to you

[@promathacademy8512]

@@jazzysocksdude we sent the solution to your email. The function is bijective!!!! 😁

[@stevencheng7997]

for the second example, shouldn't |x| > 0 but not >= due to natural number? Im still very confused on this topic I hope you could clear my confusion.. thanksss

[@promathacademy8512]

|x| >=0 for this question. Remember x comes from the domain and here the domain is Z which includes 0. The function is f(x) =|x|+2 so .........f(x) >0. |x| is a function but it is not the subject of the question by itself. Always consider what is your domain. Let me know that answers the question.

[@stevencheng7997]

@@promathacademy8512 ohhhh snap yes |x| can be >= 0 but not |x|+2 >=, is that right

[@promathacademy8512]

@@stevencheng7997 |x|+2>=2. Remember this depends on the domain of your function. I cannot emphasize this enough. For question 2 our domain was all integers but if the domain changed to "natural numbers" the result would be different.

[@dv4346]

Why did you add 10 and subtract 10 in the last question?

[@promathacademy8512]

By adding 10 and subtracting 10, this allows us to simply the expression by writing one as x divided by x. It is a manipulation technique used in algebra used to get the expression in a certain form.

[@elizabethkandjumbi3103]

At 17:07 where did the 10s come from?

[@promathacademy8512]

adding 10 and subtracting 10 is a mathematical expression which keeps the previous expression the same. Here we needed a multiple of the denominator. (a-2) so we use 5(a-2), which is 5a-10. However we only have 5a+1 but we need 5a-10 somehow, so we subtract 10 and added 10 so that 5(a-2)/(a-2) simplifies to 5 and 5(b-2)/(b-2) simplifies to 5 which simplifies the whole equation.

[@thefxbro433]

(i) Prove that the function f : [1, ∞) −→ [2, ∞) given by f(x) = (x + 2) is not surjective.(please help)

[@promathacademy8512]

The range of the function is not equal to the codomain. Simple ! Find the range of x+2

[@danielraphael919]

The range seem to be [3, ∞) while the codomain is [2, ∞) so since the range is not identical to the co domain you can conclude the the function is not surjective.

[@IKdeoSSa_7]

20:53 is my doubt

[@promathacademy8512]

So a=5? Be specific ?

[@IKdeoSSa_7]

@@promathacademy8512 My bad. I should have been more specific as you said. I am referring to why you still had (a-5) or (5-a) on the left hand side if you had already divided it on the right hand side.

[@promathacademy8512]

@@IKdeoSSa_7 yes we realize. Thank you for bringing that to our attention. The (a-5) on the left should have being canceled. We were rushing to get to the end. 😅

[@IKdeoSSa_7]

@@promathacademy8512 You are welcome👍

[@aidai_imerova]

didn't get anything

[@cocacola7535]

Were you eating candy while speaking? The sound is annoying to be honest.