Question

Let A=R*R and * be the binary operation on A defined by (a,b)*(c,d)=(a+c,b+d).Prove that * is both associative and commutative.Find the identity element for * on A.Also write the inverse element of the element (3,-5) in A.​

Answer 1

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Check the binary operation * is commutative :

We know that, * is commutative if (a, b) * (c, d) = (c, d) * (a, b) ∀ a, b, c, d ∈ R

L.H.S =(a, b) * (c, d)

=(a + c, b + d)

R. H. S = (c, d) * (a, b)

=(a + c, b + d)

Hence, L.H.S = R. H. S

Since (a, b) * (c, d) = (c, d) * (a, b) ∀ a, b, c, d ∈ R

* is commutative (a, b) * (c, d) = (a + c, b + d)

Check the binary operation * is associative :

We know that * is associative if (a, b) * ( (c, d) * (x, y) ) = ((a, b) * (c, d)) * (x, y) ∀ a, b, c, d, x, y ∈ R

L.H.S = (a, b) * ( (c, d) * (x, y) ) = (a+c+x, b+d+y)

R.H.S = ((a, b) * (c, d)) * (x, y) = (a+c+x, b+d+y)

Thus, L.H.S = R.H.S

Since (a, b) * ( (c, d) * (x, y) ) = ((a, b) * (c, d)) * (x, y) ∀ a, b, c, d, x, y ∈ N

Thus, the binary operation * is associative

Checking for Identity Element:

e is identity of * if (a, b) * e = e * (a, b) = (a, b)

where e = (x, y)

Thus, (a, b) * (x, y) = (x, y) * (a, b) = (a, b) (a + x, b + y)

= (x + a , b + y) = (a, b)

Now, (a + x, b + y) = (a, b)

Now comparing these, we get:

a+x = a

x = a -a = 0

Next compare: b +y = b

y = b-b = 0

Since A = R × R, where x and y are the natural numbers. But in this case, x and y is not a natural number. Thus, the identity element does not exist.

Therefore, the operation * does not have any identity element.

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Answer 2

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We have,

$$A=R\times R\] and \[*$$ be the binary operation on A defined by

(a,b)∗(c,d)=(a+c,b+d).

Show that

(1). ∗ is commutative

(2). ∗ is Associative

(3). Find the identity element for ∗ on A.

Then,

Proof:-

(1).∗ is commutative if

(a,b)∗(c,d)=(c,d)∗(a,b)∀a,b,c,d∈R

(a,b)∗(c,d)=(a+c,b+d)

(c,d)∗(a,b)=(c+a,d+b)

=(a+c,b+d)

Hence,

(a,b)∗(c,d)=(c,d)∗(a,b)∀a,b,c,d∈R

∗ is commutative.

 

(2). ∗ is Associative if

(a,b)∗((c,d)∗(e,f))=((a,b)∗(c,d))∗(e,f)∀a,b,c,d,e,f∈R

(a,b)∗((c,d)∗(e,f))=(a,b)∗(c+e,d+f)

=(a+c+e,b+d+f)∀a,b,c,d,e,f∈R

((a,b)∗(c,d))∗(e,f)=(a+c,b+d)∗(e,f)

=(a+c+e,b+d+f)∀a,b,c,d,e,f∈R

Since,

(a,b)∗((c,d)∗(e,f))=((a,b)∗(c,d))∗(e,f)∀a,b,c,d,e,f∈R

Hence, ∗ is associative.

 

(3).let

e is the identity element of ∗

Then,(a,b)∗e=e∗(a,b)=(a,b)

Where, e=(x,y)

So,

(a,b)∗(x,y)=(x,y)∗(a,b)=(a,b)

⇒(a+x,b+y)=(x+a,b+y)=(a,b)

No comparing that,

(a+x,b+y)=(a,b)

a+x=a,b+y=b

x = 0,y=0

Since,

A = R×R

x and y are real numbers

since 0 is real numbers.

Then identity element exist.