Relational Algebra (Natural Join Operation)

[@Ashahp11]

Hello your DBMS PLAYLIST was very much helpful for our university exams please do continue the series asap please

[@namensfindung6635]

Output:
S_ID / S_Name / Rating / Age / B_ID / Date
22 / Tom / 7 / 20 / 101 / 10-May-21
58 / Rose / 10 / 21 / 103 / 05-Jan-22

[@A-ONE991]

Yes correct

[@learningstudiesetc]

Nesco is the best ❤

[@sumeetkumar6900]

Yesc

[@anna-louisebrown8348]

Really great thank you. I didn't understand the way it was explained in my university course and you saved me!

[@RapidPodTalks]

2 rows ,6 columns

[@zohrehbakhshi-d3w]

6 attributes and 2 tuples

[@roninx9013]

correct answers please share your phone number i want to have a group session of dbms with you

[@oh-every-thing2914]

Two rows in output relation

[@mintesinotatnafu3591]

There will be only two rows only, since there is only one common attributes between the two tables, we only need to compare the values or instances, so we get 22 and 58 are present on both tables so we will have two tuples with 7 columns.

[@kevin4406]

thanks for your help! i finished this series and I found it really helpful :)

[@mr.senior5523]

S_ID,S_Name,Rating,Age,B_ID,Date

[@rajnishgupta6112]

2 tuple and 7 attribute

[@lockdowntrendings1207]

Attributes must be unique right? But sid appears two times

[@arunnautiyal2087]

109 is null

[@anonymousman2736]

6 attributes and 2 tuples..

[@AYUSHKUMARSHAW-pq9og]

suppose in natural join, if the value of ratings in case of s_id of both table are different them what should be the output.

[@ronielkamung]

2 rows❤

[@legitimate_kangaroo]

💯