Let f :X\(\to\) Y be an invertible function. Show that f has unique inverse.
(Hint: suppose g1 and g2 are two inverses of f. Then for all y∈ Y, fog1(y) = IY (y) = fog2 (y). Use one-one ness of f).
Let f :X\(\to\) Y be an invertible function. Show that f has unique inverse.
(Hint: suppose g1 and g2 are two inverses of f. Then for all y∈ Y, fog1(y) = IY (y) = fog2 (y). Use one-one ness of f).
Solution and Explanation
Let f : X → Y be an invertible function. Also, suppose f has two inverses (say ).
Then, for all y ∈Y, we have: fog1 (y)=Iy (y)=fog2 (y)
=>f(g1 (y))=f(g2 (y))
g1 (y)=g2 (y) [f is invertible =>f is one-one
=>g1 = g2 [g is one-one].
Hence, f has a unique inverse.
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- Relation R in the set A={1,2,3,...13,14} defined as R={(x,y): 3x-y=0}.
- Relation R in the set of N natural numbers defined as R={(x,y): y=x+5 and x<4}.
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R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
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Determine whether each of the following relations are reflexive, symmetric, and transitive.
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- Relation R in the set of N natural numbers defined as R={(x,y): y=x+5 and x<4}.
- Relation R in the set A={1,2,3,4,5,6} defined as R={(x,y): y is divisible by x}.
- Relation R in the set of Z integers defined as R={(x,y): x-y is an integer}
- Relation R in the set of human beings in a town at a particular time given by
- R={(x,y): x and y work at same place}
- R={(x,y): x and y live in the same locality}
- R={(x,y): x is exactly 7 am taller that y}
- R={(x,y): x is wife of y}
- R={(x,y):x is father of y}
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.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.Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive
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