Question

a binary operation * on the set R of a real number is defined by a*b =a/b+b/a where a and b belongs to R. simplify (x+1)*2=3​

Answer 1

Answer:

Solution

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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

xandyarerealnumbers

since0isrealnumbers.

Thenidentityelementexist.

Hence, this is the answer.