MATH 1302 - AY2021-T5 Unit 6 PDF

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Unit 6

PM always following this

by Saadia El-Obadi (Instructor) - Thursday, 22 July 2021, 10:01 AM

Let

with

clock.

. Discuss the properties of . Is it injective, surjective, bijective, is it a

Due dates/times displayed in activities will vary with

function? Why or why not? Under what conditions changes this?

your chosen time zone,

Explain using examples.

however you are still bound

You can use resources you nd on the internet, but use your own words and cite any resources correctly.

to the 11:55 PM GMT-5 deadline.

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Re: Unit 6 by Hebron Anyanwu - Thursday, 22 July 2021, 1:17 PM Lets begin by dening our parameters; R (Real numbers) This is a set containing all numbers in existence, both rational and irrational. Function A function is a rule or formular which assigns one element of a set to an element of another set. A function exists between to sets, the rst is called the " domain" and the second, " codomain". If an element of the domain is assigned to the codomain, then that element of the codomain is called " the

image" of the domain's element Injective function This is a function, f whereby all elements in f's codomain is the image at most one element in it's domain, i.e, an element in f's codomain is mapped or assigned to exactly one element in the domain. Surjective function This is a function, f whereby all elements in f's codomain is the image at least one element in it's domain, i.e, an element in f's codomain mustbe mapped or assigned to exactly one or more element in the domain. In this function every element in the codomain is in the range. Bijective function

This is a function, f which is both injective and surjective. Every element in the codomain must be the image of at least and at most one element in the domain, i.e, all elements in the codomain must be in the in range and be assigned to a single domain element.

In the question we were given a function whose domain and codomain are a set of real numbers; f : R→ R whose rule is; f(x) =√(x) By considering the possible elements for the codomain of f from its domain, we can conclude that; f is a function f is a function because it gives a single output for a given input. If the input is positive, e.g, 4 then we would have a single output, 2, whereas if the input is negative, e.g -4, then we have a single output, 2 i. The square root of a negative number gives a number with the "i" notation which stands for "imaginary number", this is because no square would result in a negative number. Now that we know f is a function, lets nd out if it isinjective, surjective and/or bijective. We now that no two numbers have the same square root, e.g f(0) = 0 and f(100) = 10, therefore f is injective. For a function to be surjective, it must have all the elements in it's codomain inside it's range, therefore f is not surjective because f(1) = 1 and f(1.1) =1.04880884817. Clearly there are values found in the codomain which are not in the range values. Since f is not surjective, then f is not bijective. f would be injective surjective and bijective if it had the rule; f(x) = (x). Lastly, the reason √(x)gives a single output and not ±, is because√ is asked in the question, it was a given part of the problem.√ gives a ± when we introduce it into the problem during our working (simplied radical, 2013).

Reference Simplied radical,(2013). Retrieved from https://www.youtube.com/watch?v=58giTx9R_Wg. 520 words

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Re: Unit 6 by Connor Stukes - Tuesday, 27 July 2021, 2:35 AM Hi Hebron, I liked your very detailed reply. You went step by step with the functions and the algebra to prove each type of function well. You mentioned f(x) = (x) would be a new condition that changes the function type. Good job. 43 words

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Re: Unit 6 by Hebron Anyanwu - Wednesday, 28 July 2021, 10:59 PM Hi Connor, I appreciate your response and am glad you agree the post. I wish you a success and favour in life. 22 words

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Re: Unit 6 by Irving Gonzalez Islas - Tuesday, 27 July 2021, 5:32 AM Hi Hebron Anyanwu,

Great job and thank you for the detailed response. I also came to the same end result and the easiest way to change the function is to just have it equal f(x) = f(x). What do you think would be the easiest way to write up a bijective function? 53 words

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Re: Unit 6 by Hebron Anyanwu - Wednesday, 28 July 2021, 10:57 PM Hi Irving, Thank you for your comment. I feel to write a bijective function would be to use the same rule (f(x) = x) as long as the functions domain and codomain are the same set, for example, f : Z → Z. If the two aren't the same then the rule would depend heavily on the codomain, for instance, the codomain would have to have exactly the same cardinality as the domain. I hope this response was to your liking. Thank you and I wish you good luck and success. 89 words

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Re: Unit 6 by Nam Nguyen - Tuesday, 27 July 2021, 2:05 PM Hi Hebron,

The explanation is very well explained. Well done.

Best Regards.

Nam 13 words

Re: Unit 6 by Hebron Anyanwu - Wednesday, 28 July 2021, 11:00 PM Hi Nam, Thank you for your response, I also wish you my best regards. 14 words

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Re: Unit 6 by Hiroyuki Kitagawa - Tuesday, 27 July 2021, 5:50 PM Hi Hebron, Very nice part introducing all the denitions. Thank you, Hiroyuki 12 words

Re: Unit 6 by Hebron Anyanwu - Wednesday, 28 July 2021, 11:02 PM Hi Hiroyuki, Your comment is very encouraging, thank you. I wish you success in your endeavors. 16 words

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Re: Unit 6 by Iliya Saleh - Wednesday, 28 July 2021, 12:21 AM Hello Hebron, I enjoyed your detailed explanation. I guess I missed the square root of a negative number which result to a complex number. Thanks again and well done. 29 words

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Re: Unit 6 by Hebron Anyanwu - Wednesday, 28 July 2021, 11:04 PM HI Iliya, I am glad you enjoyed the explanation, your response denitely made it worth the time. I wish you good luck and success in life. 26 words

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Re: Unit 6 by NEMETCHA TOWO FRANCOIS MODESTE - Thursday, 29 July 2021, 2:36 AM hi Hebron, I really appreciated your approach, I was able to learn a lot of things, I can only wish you a lot of courage and a lot of success. 30 words

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Re: Unit 6 by Yosef Tegegne - Friday, 23 July 2021, 5:24 PM

The domain and the codomain of

with

are real numbers. From our high

school algebra class, we remember that Real numbers are dened as a union of rational and irrational numbers include both negative and positive numbers. Real numbers therefore include natural numbers, decimals, fractions, integers, natural number. Real numbers do not include imaginary or complex numbers. Plotting the function  element of real numbers.

 shows us that this function is dened only for x ≥ 0 where x is an

According to Levin O. (2019), a function is dened as “a rule that assigns each element of a set, called thedomain, to exactly one element of a second set, called thecodomain.” In the function

, the domain is a real number and the codomain is also a real number.

However, it is some element of the domain f cannot be mapped to its codomain. Because of this reason, with

Therefore

fails to satisfy the property to be a function.

with

is not a function. Since it is not function, it is not injective,

surjective or bijective.

References: Levin, O. (2019).Discrete mathematics: An open introduction (3 edition). CreateSpace Independent Publishing Platform.http://discrete.openmathbooks.org/pdfs/dmoi-tablet.pdf

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Re: Unit 6 by Saadia El-Obadi (Instructor) - Monday, 26 July 2021, 1:57 PM Excellent Yosef!

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Good study of the function! 7 words

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Re: Unit 6 by Juan Aguilar - Tuesday, 27 July 2021, 3:54 AM Hello Yosef,

Thanks for your post. It helps me to understand even more the concept of functions. Your examples are well dened. You are demonstrating every single point. 28 words

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Re: Unit 6 by Irving Gonzalez Islas - Tuesday, 27 July 2021, 5:39 AM Hi Yosef Tegegne,

Great post! You have a mastery of the subject and you explained the topic in a way that help me understand the topic a little better. The I had assumed that the function was a proper function but your statement "However, it is some element of the domain f cannot be mapped to its codomain.". However I thought the function to be injective since each x can only be mapped to a certain y, thus making injective. Your post has made me realize I will need to perform further research on the topic at hand. 98 words

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Re: Unit 6 by Nam Nguyen - Tuesday, 27 July 2021, 2:05 PM Yosef,

Very well explained. Good work

Nam 7 words

Re: Unit 6 by Ayman Emam - Tuesday, 27 July 2021, 6:20 PM good work and great example 5 words

Re: Unit 6 by Pratik Timalsina - Tuesday, 27 July 2021, 6:25 PM Dear Yosef, Great work. You've provided a detailed answer of the mentioned question with gure to understand as well. It was convenient to go through your writings. Thank you. Pratik 30 words

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Re: Unit 6 by Iliya Saleh - Wednesday, 28 July 2021, 12:44 AM Thanks Yosef, your explanation is clear and detailed enough. Thanks for reminding us that the set of real number does not include complex numbers and for the f given for set of real number it does not meet the condition as a function. Thanks 44 words

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Re: Unit 6 by Ahmed Adnew - Wednesday, 28 July 2021, 10:56 PM Hello Yosef I appriciate your condence to deside easily about the question. Actually you are right all the thing is nished at the start becouse the domain is required to judge about the function. But if you limit the domain in interval we can creat other things.so the given equestion is function for positive domain and codomain. I wish you all the best 63 words

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Re: Unit 6 by Hebron Anyanwu - Wednesday, 28 July 2021, 11:13 PM Hi Yosef, Your post is really enlightening, that little factor of real numbers not including imaginary numbers completely changes the answer anyone would derive. Thank you for your amazing post, I wish you good luck and success. 37 words

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Re: Unit 6 by Vincent Obunga - Thursday, 29 July 2021, 12:00 AM Yosef,

your explanation is informative and straight forward. Especially the interrupt concept. Nice work. 14 words

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Re: Unit 6 by Louis Malale - Thursday, 29 July 2021, 12:34 AM Hi Yosef, Well done on your post. 7 words

Re: Unit 6 by NEMETCHA TOWO FRANCOIS MODESTE - Thursday, 29 July 2021, 2:37 AM Hi Yossef, You did a good research, I read with a lot of interest I wish you a lot of courage and good luck 24 words

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Re: Unit 6 by Deborah Barbier - Thursday, 29 July 2021, 8:11 AM Hi Yosef - Thank you for this post. It has helped me to understand this topic better as I struggled with it this week. Looking forward to next week's post. 29 words

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Re: Unit 6 by Kathrine Williams - Saturday, 24 July 2021, 9:24 PM f(x) is a continuous function that produces values of R+ I believe that f: R ->R suggests that we invalidate any x value that would produce a non-real number.

Injective: “if every element of the co-domain is the image of at most one element from the domain” (Levin, 2019). Any value of the co-domain that can be a function of the square root of x can only be produced by one value of x.

Thus, the function with is Injective.

Surjective: “if every element of the co-domain is the image of at least one element from the domain” (Levin, 2019). Because the co-domain consists of all real numbers, this includes negative numbers. There is no real number that I would be able to take the square root from in order to achieve a negative number, thus not every y-value has an associated x-value.

The square root x will always be a positive value even when very small sqrt(49) = 7 sqrt(12) = 3.464 sqrt(3) = 1.732 sqrt(.22) = .469 sqrt(.0000005) = 7.07106781×10−4 which is a very small number but still not negative

Thus, the function with not Surjective.

In order for a function to be bijective, it must be both injective and surjective, which it is not. I believe that by simply changing f: R → R to R+ → R+ we eliminate all negative values, making our function both injective and surjective and this bijective.

Resources

Levin, O. (2019). Discrete mathematics: An open introduction (3 edition). CreateSpace Independent Publishing Platform. http://discrete.openmathbooks.org/pdfs/dmoi-tablet.pdf 259 words

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Re: Unit 6 by Saadia El-Obadi (Instructor) - Sunday, 25 July 2021, 8:45 PM Excellent Kathrine! Good explanation! Well done! 6 words

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Re: Unit 6 by Connor Stukes - Tuesday, 27 July 2021, 2:41 AM Hi Kathrine, I read through your reply, and it is very organized and easy to understand. I came up with the same conclusions that you did. The conditions that would change it I used were all natural numbers, which means all positive numbers, and I see that you did the same with all positive real numbers. Good job. 58 words

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Re: Unit 6 by Juan Aguilar - Tuesday, 27 July 2021, 3:56 AM Hello Kathrine,

You have a very good example and you also show dierent values for x which is a good idea to extend your point. 25 words

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Re: Unit 6 by Irving Gonzalez Islas - Tuesday, 27 July 2021, 5:47 AM Hi Kathrine Williams,

Great discussion post. I have also come to the same conclusion but another student posted an interesting theory that the function is not really a function. I wasn't sure if this was accurate but it does force me to ask if the function we are discussion is a function. I believe it is for a certain range of values. Since the function we are discussing has a squared value then we are limited to positive numbers. So thank you for a great discussion and helping me conrm my own assumptions about the topic. 96 words

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Re: Unit 6 by Hiroyuki Kitagawa - Tuesday, 27 July 2021, 5:53 PM Hi Katherine, You gave very nice practical example. Thank you Hiroyuki 11 words

Re: Unit 6 by Ayman Emam - Tuesday, 27 July 2021, 6:20 PM

great explanation 2 words

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Re: Unit 6 by Pratik Timalsina - Tuesday, 27 July 2021, 6:27 PM Dear Kathrine, Everything is described clearly regarding the mentioned question. It was clear and convenient to go through your writings. Overall, great work. Thank you. Pratik 26 words

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Re: Unit 6 by Iliya Saleh - Wednesday, 28 July 2021, 1:02 AM Hello Kathrine, your explanation is clear and well Understood. It is clear that the set of real number, you can not nd the square root of a negative value, which implies that the negative values in the domain are not assigned to any value in the codomain which means f is not a function. Thanks for your explanation, and well done. 61 words

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Re: Unit 6 by Yosef Tegegne - Wednesday, 28 July 2021, 5:33 PM Hi Kaherine,

Your analysis is wonderful and I fully agree with your nal conclusion. Thank you for your handwork. 19 words

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Re: Unit 6 by Goran Karlic - Wednesday, 28 July 2021, 9:21 PM Dear Kathrine,

an excellent post that succinctly and correctly explains each concept as it relates to the function in question.

Cheers, Goran 22 words

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Re: Unit 6 by Hebron Anyanwu - Wednesday, 28 July 2021, 11:17 PM Hi Katherine, Lovely post lled with amazing points but I am condent that this is not a function due to the lack of possible values for any of the negative numbers in the domain. 34 words

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Re: Unit 6 by Vincent Obunga - Thursday, 29 July 2021, 12:01 AM Kathrine, Your explanation is concise and straight to the point. Thanks to your post, my understanding is much clearer now. Well written and great work! 25 words

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Re: Unit 6 by NEMETCHA TOWO FRANCOIS MODESTE - Thursday, 29 July 2021, 2:38 AM hi kathrine, I read with great interest your contribution of this week, I can only wish you much success for the futur 22 words

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Re: Unit 6 by Deborah Barbier - Thursday, 29 July 2021, 8:12 AM Hi Katherine - Your examples helped me to understand this topic in depth. Thank you for your discussion. Looking forward to next week's. 22 words

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Re: Unit 6 by Connor Stukes - Saturday, 24 July 2021, 10:43 PM

To determine if f(x) = sqrt(x) with f: R->R is injective, surjective, bijective or a function, you have to test each one separately (Math is Fun, n.d.).

To determine if it is injective: Given: x1 and x2 are real numbers. If f(x1) = f(x2), then it is injective if both have a one to one A to B point. 2

2

(sqrt x1 ) = (sqrt x2 ) , square both sides, then x1 = x2 Since squaring both sides results in one value for x 2, then we have a one to one A to B point. Therefore, it is injective and a function. To determine if it is surjective: Given: We proved f(x) is a function above with all real numbers. It is surjective when every B in the function has an A in the function. Test a negative number like -4. If f(x) = -4, the the square root of -4 is not a real number (it is imaginary). This means not every B has an A. Ther...