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1.

Let * be a binary operation on set `Q-[1]`defined by `a*b=a+b-a b`for all`a , b in Q-[1]dot`Find the identity element with respect to `*onQdot`Also, prove that every element of `Q-[1]`is invertible.

Answer» Let `e` is the identity element for the given operation.
Then, `a**e = a`
`=>a+e-ae = a`
`=> e(1-a) = 0`
`=> e = 0 and a = 1`
It is given that `Q` does not contain `1`.
`:. a` can not be `1`.
`:. e = 0`
So, identity element for the given operation is `0`.
Now, we will find the invertible elements of `Q-[1]`.
Let `x` is the inverse of `a`
Then, `a**x = e`
`=>a+x-ax = 0`
`=>x(1-a) = -a`
`=>x = a/(a-1)`
So, we will not be able to find the inverse if `a = 0`.
For, all other values of `a`, we can find the inverse.
So, all elements of `Q-[1]` are invertible.
Discussion
2.

Let `A=Q x Q`and let * be a binary operation on A defined by`(a , b)*(c , d)=(a c , b+a d)`for `(a , b),(c , d) in Adot`Then, with respect to * on AFind the identity element in AFind the invertible elements of A.

Answer» An identity element `e` in a relation is an element such that
`a**e = e**a = a.`
Here, `(a,b)**(c,d) = (ac,b+ad)`
Let `(c,d)` is the identity element for the given operation.
Then, `ac = a and b+ad = b`
`=>c = 1 and ad = 0`
`=>c = 1 and d = 0`
So, identity element for the given operation is `(1,0)`.
Now, we will find the invertible elements of `A`.
Let `(x,y)` are the invertible elements of `A`.
Then,
`(a,b)**(x,y) = (1,0)`
`=>ax = 1 and b +ay = 0`
`=>x = 1/a and y = -b/a`
So, invertible elements of `A` will be in form of `(1/a,-b/a).`
Discussion
3.

On Q, the set of all rational numbers, a binary operation * is definedby `a*b=(a b)/5`for all `a , b in Qdot`Find the identity element for * in Q. Also, prove that every non-zeroelement of Q is invertible.

Answer» Here, `a**b = (ab)/5`.
An identity element `e` in a relation is an element such that
`a**e = e**a = a.`
So, in the given relation,
`a**e = (ae)/5 = a => ae = 5a => e = 5`
So, identity element is `5` for the given relation.
Now, for any element `x in Q`,
If `a**x = e`, then `x` is inverse of `a`.
Here, `a**x = 5 => (ax)/5 = 5`
`x = 25/a`
So, inverse of given relation is `25/a` where `a in Q`.
So, `Q` is invertible for every non-zero element.
Discussion
4.

Define a binary operation * on the set `A={1,2,3,4}`as `a*b=a b`(mod 5). Show that 1 is the identity for * andall elements of the set A are invertible with`2^(-1)=3`and `4^(-1)=4`

Answer» Here, `A = {1,2,3,4}.`
And, `a**b = ab mod 5.`So, if we take any two elements from `A`,
Then we can define the given binary operation.
For example,`1**1 = (1*1) mod 5 = 2`
`2**2 = (2*2) mod 5 = 4`
`2**3 = (2*3) mod 5 = 1`
`4**4 = (4*4) mod 5 = 1`
Similarly, we can define the given binary operation on each order pair of `A`.
Please refer to video to see the complete table for this binary operation.
Now, we will find the identity element of this relation.
An identity element `e` in a relation is an element such that
`a**e = e**a = a.`
In this case, identity element is `1`.
`:. e = 1.`
Now, we know, if `ab = e`, then `a = b^-1`.
Here, `2**3 = 1 => 2^-1 = 3`.
Also, `4**4 = 1 => 4^-1 = 4`.
Discussion
5.

If * is defined on the set R of all real numbers by `a*b=sqrt(a^2+b^2)`, find the identity element in R with respect to *.

Answer» An identity element `e` in a relation is an element such that
`a**e = e**a = a.`
Here, `a**e = sqrt(a^2+e^2)`
`:. sqrt(a^2+e^2) = a`
Squaring both sides,
`=>a^2+e^2 = a^2`
`=>e^2 = 0`
`=>e = 0`
So, identity element for the given relation is `0`.
Discussion
6.

Let S be the set of all rational numbers except 1 and * be defined on Sby `a*b=a+b-a b ,`for all`a , b in Sdot` Find its identity element

Answer» Let `e` is the identity element for the given operation.
Then, `a**e = a`
`=>a+e-ae = a`
`=> e(1-a) = 0`
`=> e = 0 and a = 1`
It is given that `S` does not contain `1`.
`:. a` can not be `1`.
`:. e = 0`
So, identity element for the given operation is `0`.
Discussion