Relations and Functions
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New answer posted
5 months agoContributor-Level 10
The binary operation? on the set {1, 2, 3, 4, 5} is defined as a? b = min {a, b} &mn For E; a, b? {1, 2, 3, 4, 5}.
Thus, the operation table for the given operation? can be given as:
? | 1 | 2 | 3 | 4 | 5 |
1 | 1 | 1 | 1 | 1 | 1 |
2 | 1 | 2 | 2 | 2 | 2 |
3 | 1 | 2 | 3 | 3 | 3 |
4 | 1 | 2 | 3 | 4 | 4 |
5 | 1 | 2 | 3 | 4 | 5 |
New answer posted
5 months agoContributor-Level 10
(i) On Z, * is defined by a * b = a − b.
It can be observed that 1 * 2 = 1 − 2 = 1 and 2 * 1 = 2 − 1 = 1.
∴1 * 2 ≠ 2 * 1; where 1, 2 ∈ Z
Hence, the operation * is not commutative.
Also we have:
(1 * 2) * 3 = (1 − 2) * 3 = −1 * 3 = −1 − 3 = −4
1 * (2 * 3) = 1 * (2 − 3) = 1 * −1 = 1 − (−1) = 2
∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Z
Hence, the operation * is not associative.
(ii) On Q, * is defined by a * b = ab + 1.
It is known that:
ab = ba & mn For E; a, b ∈ Q
⇒ ab + 1 = ba&nb
New answer posted
5 months agoContributor-Level 10
(i) On Z+, * is defined by a * b = a − b.
It is not a binary operation as the image of (1, 2) under * is 1 * 2 = 1 − 2= −1 ∉ Z+.
(ii) On Z+, * is defined by a * b = ab.
It is seen that for each a, b ∈ Z+, there is a unique element ab in Z+.
This means that * carries each pair (a, b) to a unique element a * b = ab in Z+.
Therefore, * is a binary operation.
(iii) On R, * is defined by a * b = ab2.
It is seen that for each a, b ∈ R, there is a unique element ab2 in R.
This means that * carries each pair (a, b) to a unique element a * b = ab2 in R.
Therefore, * is a binary operation.
(iv) On Z+, * is defined by a * b = |a −
New answer posted
5 months agoContributor-Level 10
It is given that is defined as
Let y be an arbitrary element of Range f.
Then, there exists such that
let us define g: Range
Thus, g is the inverse of f i.e.,
Hence, the inverse of f is the map g: Range which is given by
The correct answer is B.
New answer posted
5 months agoContributor-Level 10
Let f: X → Y be an invertible function.
Then, there exists a function g: Y → X such that gof = IX and fog = IY.
Here, f−1 = g.
Now, gof = IX and fog = IY
⇒ f−1 of = IX and fof−1 = IY
Hence, f−1 : Y → X is invertible and f is the inverse of f−1
i.e., (f−1)−1 = f.
New answer posted
5 months agoContributor-Level 10
If we define then we have:
Thus, the inverse of f exists and
is given by,
Let us now find the inverse of i.e., find the inverse of g.
If we define
, then we have
Thus, the inverse of g exists and
It can be noted that h=f.
Hence,
New answer posted
5 months agoContributor-Level 10
Let be an invertible function.
Also, suppose f has two inverses (say g1 and g2 ).
Then, for all y ∈ Y, we have:
[f is invertible => f is one-one]
[g is one-one]
Hence, f has a unique inverse.
New answer posted
5 months agoContributor-Level 10
is given by,
f is a one-one function.
Onto:
Therefore, for any , there exists such that
f is onto.
Thus, f is one-one and onto and therefore, exists.
Let us define by
Hence, f is invertible and the inverse of f is given by
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