Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.

Answer
It is given that A = {1, 2, 3}, B = {4, 5, 6, 7}.
f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}.
∴ f (1) = 4, f (2) = 5, f (3) = 6
It is seen that the images of distinct elements of A under f are distinct.
Hence, function f is one-one.

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Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.

Answer 

Method -1 

Let A = {1, 2, 3, 4, 5, 6}.

A relation R is defined on set A as: R = {(a, b): b = a + 1}

∴R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} We can find (a, a) ∉ R, where a ∈ A.

For instance,

(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∉ R

∴R is not reflexive.

It can be observed that (1, 2) ∈ R, but (2, 1) ∉ R.

∴R is not symmetric. Now, (1, 2), (2, 3) ∈ R

But,

(1, 3) ∉ R

∴R is not transitive

Hence, R is neither reflexive, nor symmetric, nor transitive.

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Method- 2 

 

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Do you know any other method of solving of this question ?

3. Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.

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