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Showing posts with label 2. Show all posts
Showing posts with label 2. Show all posts
Thursday, 20 June 2013
Tuesday, 11 June 2013
Sunday, 9 June 2013
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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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.
You can Type your comment on question 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. below box
Labels:
(2,
(3,
2,
3},
4),
5,
5),
6,
6)} be a function from A to B. Show that f is one-one.,
7} and let f = {(1,
B = {4,
Let A = {1
Saturday, 8 June 2013
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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