0 votes
99 views
in Chapter 1 Relations and Functions by (8.1k points)
edited
Show that none of the operations given above has identity.

1 Answer

0 votes
by (8.1k points)
selected by
 
Best answer
The binary operation * on set Q is
(i) defined as a * b = a – b.
For idenity element e, a * e = e * a = a.
But a * e = a – e # a and e * a = e – a # a.
∴ There is no identity element for this operation.

(ii) Binary operation * is defined as
a * b = a² + b²
Here, a * e = a² + e² # a
and e * a – e² + a² # a.
So, this operation * has no identity.

(iii) The binary operation * is defined as
a * b = a + ab
We have: a * e = a + ae # a
and e * a = e + ea # a.
∴ There is no identity element.

(iv) The binary operation * is defined as
a * b = (a – b)²
Put b = e. We get a * e = (a – e)² # a
and e * a = (e – a)² # a
for any value of e ∈ Q.
⇒ There is no identity element.

(v) The operation * is
a * b = \(\frac { ab }{ 4 }\).
and e * a = \(\frac { ab }{ 4 }\) # a
for any value of e ∈ Q.
∴ Operation * has no identity element.

(vi) The operation * is a * b = ab².
Put b = e. We get a * e = ae² # a
and e * a = ea² # a
for any value of e ∈Q.
⇒ There is no identity element.
Thus, these operations have no identity.

Related questions

Doubtly is an online community for engineering students, offering:

  • Free viva questions PDFs
  • Previous year question papers (PYQs)
  • Academic doubt solutions
  • Expert-guided solutions

Get the pro version for free by logging in!

5.7k questions

5.1k answers

108 comments

559 users

...