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Relations and Functions

Short Answer Type

311. Show that the number of binary operations on {1, 2} having 1 as identity and having 2 as the inverse of 2 is exactly one.

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312.

State whether the following statements are true or false. Justify.

(i) For an arbitrary binary operation * on a set N, a*a = a ∀ a ∈N.
(ii) If * is a commutative binary operation on N, then a * (b * c) = (c * b) * a.

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313.

Consider a binary operation * on N defined as a * b = a³+ b³ . Choose the correct answer.

(A)    Is * both associative and commutative ?
(B)    Is * commutative but not associative?
(C)    Is * associative but not commutative?
(D)    Is * neither commutative nor associative?

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314. Number of binary operations on the set {a, b} are (A) 10 (B) 16 (C) 20 (D) 8

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315.

Let A = Q x Q. Let be a binary operation on A defined by (a, b) * (c, d) = (ac, ad + b).

Then

(i) Find the identify element of (A, *)
(ii) Find the invertible elements of (A, *)

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316.

Define binary operation ‘*’ on Q as follows : a * b = a + b – ab, a, b∈ Q
(i) Find the identity element of (Q, *).
(ii) Which elements in (Q. *) are invertible?

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317. Show that – a is the inverse of a for the addition operation ‘+’ on R and

1 over straight a

is the inverse of a ≠ 0 for the multiplication operation ‘x’ on R.

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318. Show that addition, subtraction and multiplication are binary operations on R, but division is not a binary operation on R. Further, show that division is a binary operation on the set R. of non-zero real numbers.

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319. Show that f : N → N, given by

$straight f left parenthesis straight x right parenthesis space open curly brackets table row cell straight x space space plus end cell 1 row cell straight x space minus end cell 1 end table table row if row if end table space table row cell straight x space is space odd end cell row cell straight x space is space even end cell end table close curly brackets$

Let x, m be any two odd natural numbers.

Then f(x) = f (m) ⇒ x + 1 = m + 1 ⇒ x = m Let x, m be any two even natural numbers.

Then f (x) = f (m) ⇒ x – 1 = m – 1 ⇒ x = m

is both the cases, f (x) = f(m) ⇒ x = m If x is even and m is odd, then x ≠ m Also f (x) is odd and f (m) is even. So, f (x) ≠ f(m) ∴ x ≠ m ⇒ f (x) ≠ f (m) ∴ f is one-to-one.

Now, let x be an arbitrary natural number. If x is an odd number, then there exists an even natural number (x + 1) such that f(x + 1) = x + 1–1 = x. If x is an even number, then there exists an odd natural number (x – 1) such that

$straight f left parenthesis straight x right parenthesis space open curly brackets table row cell straight x space space plus end cell 1 row cell straight x space minus end cell 1 end table table row if row if end table space table row cell straight x space is space odd end cell row cell straight x space is space even end cell end table close curly brackets$

Let x, m be any two odd natural numbers.
Then f(x) = f (m) ⇒ x + 1 = m + 1 ⇒ x = m Let x, m be any two even natural numbers.
Then f (x) = f (m) ⇒ x – 1 = m – 1 ⇒ x = m
is both the cases, f (x) = f(m) ⇒ x = m If x is even and m is odd, then x ≠ m Also f (x) is odd and f (m) is even. So, f (x) ≠ f(m) ∴ x ≠ m ⇒ f (x) ≠ f (m) ∴ f is one-to-one.
Now, let x be an arbitrary natural number. If x is an odd number, then there exists an even natural number (x + 1) such that f(x + 1) = x + 1–1 = x. If x is an even number, then there exists an odd natural number (x – 1) such that
f (x – 1) = (x – 1) + 1 = x
∴ every x ∈ N has its pre-image in N
∴ f : N → N is onto.
∴ f is one-to-one and onto.

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320. Construct a 3 X 2 matrix whose elements are given by

straight a subscript straight i space straight j end subscript equals 1 half space left square bracket space straight i minus 3 space straight j space straight i.

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