Relational Algebra (Select Operation)

[@ismailfaridi]

Solution: 1. σ Salary > 60000 (INSTRUCTOR) 2. σ Salary ≥ 50000 ^ Salary ≤ 75000 (INSTRUCTOR) 3. σ 1=1 (INSTRUCTOR)

[@HugeCraziness]

Excellent, The relational algebra expression σ 1=1 (INSTRUCTOR) is equivalent to σ True(INSTRUCTOR). Since the condition 1=1 is always true, the Selection operator returns all tuples in the INSTRUCTOR table 1.

[@nigelthornberry007]

@@HugeCraziness Thank you for the explanation!!

[@VimuThe5ifer]

in the second question, what if σ Salary > 50000 ^ Salary < 75000 (INSTRUCTOR) it's between 50000 and 75000, does it need to be "Salary ≥ 50000 ^ Salary ≤ 75000"🤔 Tell me if I'm wrong

[@kharris274]

@@VimuThe5ifer then the "Salary = 50000" and "Salary = 75000" won't be included

[@VimuThe5ifer]

@@kharris274 yeah I know, but in the question it doesn't mention that it should include 50000 and 75000. it says 'salary between 50K and 75K'. If you got this question for an exam or something, What should be the correct answer🤔🤔 these kind of questions confuses me

[@HugeCraziness]

The relational algebra expression σ 1=1 (INSTRUCTOR) is equivalent to σ True(INSTRUCTOR). Since the condition 1=1 is always true, the Selection operator returns all tuples in the INSTRUCTOR table 1.

[@vaaipuillaraja3131]

σ 1=1 (INSTRUCTOR) how?

[@YousefRazzouk-ve1bv]

@@vaaipuillaraja3131 we deal with the subscript as a boolean condition whether it was about attributes in the relation or any other thing

[@vaaipuillaraja3131]

@@YousefRazzouk-ve1bv Thank you

[@VedangRaul]

1. sigma salary>60000(INSTRUCTOR) 2. sigma salary >=50000 AND salary 1(INSTRUCTOR)

[@AosrtyAosrty]

select Salary > 60000 (Instructor) , select Salary >= 50000 ^ Salary = 1 (Instructor)

[@hecticgamings]

Wrong 1st me gretaer than equL to nhi likha h then why u written here >=

[@hecticgamings]

Greater than means use bada naki uske equL

[@archanavr8974]

1:39 1:39 1:39

[@NaveenSaiCremsiYadlapalli]

1) σ Salary>60000(INSTRUCTOR) 2) σ Salary>=50000 ^ Salary

[@reviver_z]

I was preparing all summer for these exams . I hope it goes well .

[@sttaseen5106]

For q3, you can just use a tautology (a condition that is always true) instead of using id. It would look something like this: select 1=1 (INSTRUCTOR). Using id is a bit finicky because we don't know if it is a string, a number and if it is, can it be negative etc.

[@kishanpayadi8168]

Noice.

[@jonanderssunde]

Nice answer for Q3

[@ayamwenzdayz3551]

Oh nice

[@iamsherk2605]

youre a better instructor than my instructor

[@reverbism]

1. SELECT Salary > 60,000 (Instructor) 2. SELECT Salary > 50,000 AND Salary < 75,000 (Instructor) 3. SELECT ID > 0 (Instructor) 3. SELECT (Instructor) //Not work coz no condition here

[@hbfl3x50]

for no.3 we can also use this long way which i think is easy to understand but exhausting SELECT Dept_Name = "Biology" OR Dept_Name = "Computer Science" OR Dept_Name = "Electrical" OR Dept_Name = "Finance" OR Dept_Name = "Music" OR Dept_Name = "History".

[@Ena_aysha]

can t we use - select true (instractor)

[@clipstv6229]

15:51 Q3 σ ID=¬NULL (INSTRUCTOR) is this correct?

[@aimanmehar9580]

Q3: select id≠0 (INSTRUCTOR)

[@feelthemusic5677]

awesome 🤩

[@jeevar5424]

1.Salary>60000(INSTRUCTOR) 2.50000

[@AbhishekLamba-yd3dq]

question 3 condition is id != null

[@xMichi7]

Thank you!

[@rheniuspaul389]

Please cover dbms quicker so that it can be useful for exams

[@jokishwieu1641]

Thanks a lot, this video has incredibly helped me a lot, you're legend man

[@katharinast06]

you are amazing. thank you so much!

[@PraisySanthaMalarP]

σ TRUE(INSTRUCTOR) is this correct for ques 3

[@kumarakalvamanjunath8353]

Superb class 👏👍

[@diegoortega2374]

My solution for the last one: σ name != null (INSTRUCTOR)

[@shailendra_25]

Superb Explanation 👌

[@patronerox]

great work, butit would be awesome if you provided footprints of every course in pdf

[@sophanithphon]

I don't understand question number 3.

[@SHIVESHKUMAR-q7i]

INSTRUCTOR OR Sigma ID > 0(INSTRUCTOR)

[@olukayodepaul7070]

σ salary>60000 (INSTRUCTOR) 2. σ salary >=50000 ^ salary 1 (INSTRUCTOR)

[@vishalsisodiya28]

Can I get fuel?

[@ernestkeya-sq4sh]

What if someone talks about solving using sql is it tge same thing?

[@custompixel3364]

select ID not equal to zero(Instructor) because ID never can be zero.

[@theamazingtutorial.9019]

ID is always in string. So we can not compare it with an integer value

[@kamenpetkov8179]

I would argue that for question 3 you can simply write INSTRUCTOR (just the relation basically) which would output all of the rows and columns in the relation. I don't think it's necessary to use Selection for this one.

[@jaganaanusha2001]

Super explanation 👍

[@ENagamani-yl1ub]

1.sigma salary>60000(INSTRUCTOR) 2.sigma salary=50000 or salary=75000(INSTRUCTOR) 3.sigma id>10000(INSTRUCTOR)

[@v.balajisai424]

Superb ✌️

[@yabon87]

1. σ Salary >60000(INSTRUCTOR) 2. σ Salary>=50000 AND Salary

[@rayyao9941]

you are my legend

[@jilyanzmeigh9132]

THANK U SM

[@ummeed8821]

σ true (INSTRUCTOR)

[@reviver_z]

Because i'm confused.. How can we describe this SQL to Relational algebra? SELECT a.name,a.surname,b.LessonCode FROM student AS a LEFT JOIN StudentInLesson AS b ON a.ID=b.StudentID WHERE b.LessonCode=104;

[@pibob7880]

I wish it explained what relations are as well. so far, I know tuples, and tables are related to relations. In database theory, a relation, as originally defined by E. F. Codd,[1] is a set of tuples (d1, d2, ..., dn), where each element dj is a member of Dj, a data domain a tuple is a set of attribute values in which no two distinct elements have the same name In SQL, a database language for relational databases, relations are represented by tables, where each row of a table represents a single tuple, and where the values of each attribute form a column.

[@frederickteye]

1. Select Salary > 60000 (INSTRUCTOR) 2. Select Salary > 50000 ^ Salary < 75000(INSTRUCTOR) 3. Select ID > 1 (INSTRUCTOR)

[@sarveshkilje405]

Select = salary > 60000(INSTRUCTOR) Q2= Select = salary >50000 and salary

[@tubakzgn1102]

I did the same thing.

[@Patrick-hb9dbj]

how about Q3 dude?

[@Patrick-hb9dbj]

what can we write in the condition if we need to select all entries in a relation..???

[@pravindhotre6405]

Select= ID>0 ( INSTRUCTOR)……..3rd question ans😊

[@Patrick-hb9dbj]

@@pravindhotre6405 many thanks pravin!