
In a group \[G = \{ 1,2,3,4,5,6\} \]under \[{ \otimes _7}\], then solution \[4{ \otimes _7}x = 5\] is
a. 3
b. 2
c. 4
d. 5
Answer
528.9k+ views
Hint: Here in this question the question is related to the group and they have given binary operation. The binary operation is the multiplication modulo 7. First we write the table for the given group and the binary operation and then we determine the value of x which satisfies the given condition or the equation.
Complete step by step solution:
In mathematics, a group is a set equipped with a binary operation that combines any two elements to form a third element in such a way that three conditions called group axioms are satisfied, namely associativity, identity and invertibility. If the conditions satisfy then the set will form a group.
Using the binary operation we will write the table for the given set
Here we write the table based on the binary operation which is a multiplication modulo 7, if the product of two numbers which is in the set exceeds the number 6 then we divide the number by 7 and then we write the remainder in the boxes of the table.
Now in the question they have given \[4{ \otimes _7}x = 5\], when we see the table. The 5th row and 3rd column gives the answer. therefore the value of x is 3
Therefore the option a is the correct one.
So, the correct answer is “Option a”.
Note: In the group we use four kinds of binary operations namely, addition, multiplication, addition modulo, multiplication modulo. The modulo term is different when we write the remainder in the table. The table is more important in the group concept. The table we write depends on the binary operation.
Complete step by step solution:
In mathematics, a group is a set equipped with a binary operation that combines any two elements to form a third element in such a way that three conditions called group axioms are satisfied, namely associativity, identity and invertibility. If the conditions satisfy then the set will form a group.
Using the binary operation we will write the table for the given set
| \[{ \otimes _7}\] | 1 | 2 | 3 | 4 | 5 | 6 |
| 1 | 1 | 2 | 3 | 4 | 5 | 6 |
| 2 | 2 | 4 | 6 | 1 | 3 | 5 |
| 3 | 3 | 6 | 2 | 5 | 1 | 4 |
| 4 | 4 | 1 | 5 | 2 | 6 | 3 |
| 5 | 5 | 3 | 1 | 6 | 4 | 2 |
| 6 | 6 | 5 | 4 | 3 | 2 | 1 |
Here we write the table based on the binary operation which is a multiplication modulo 7, if the product of two numbers which is in the set exceeds the number 6 then we divide the number by 7 and then we write the remainder in the boxes of the table.
Now in the question they have given \[4{ \otimes _7}x = 5\], when we see the table. The 5th row and 3rd column gives the answer. therefore the value of x is 3
Therefore the option a is the correct one.
So, the correct answer is “Option a”.
Note: In the group we use four kinds of binary operations namely, addition, multiplication, addition modulo, multiplication modulo. The modulo term is different when we write the remainder in the table. The table is more important in the group concept. The table we write depends on the binary operation.
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