Let f,g and h be functions from R to R. Show that (f+g)oh=foh+goh.(f.g)oh=(foh).(goh)
Solution and Explanation
To prove:(f+g)oh=foh+goh
consider:((f+g)oh)(x)=(f+g)(h(x))=(foh)(x)+(goh)(x) {(foh)+(goh)(x)}
therefore ((f+g)oh)(x)={(foh)+(goh)(x)} ∀ x ∈ R.
Hence (f+g)oh=foh+goh
To prove:(f.g)oh=(foh).(goh)
consider:((f.g)oh)(x)=(f.g)(h(x))
=(foh)(x).(goh)(x)
={(foh).(goh)}(x)
therefore ((f.g)oh)(x)={(foh).(goh)}(x) ∀ x ∈ R
Hence,(f.g)oh=(foh).(goh)
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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={(x,y):x is father of y}
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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.
- 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}.
- 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
but not symmetric.Check whether the relation R in R defined as R = {(a, b): a ≤ b3} is reflexive, symmetric or transitive
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