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Question
Prove that the function f : N → N, defined by f(x) = x2 + x + 1, is one-one but not onto
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Solution
f : N → N, defined by f(x) = x2 + x + 1
Injectivity:
Let x and y be any two elements in the domain (N), such that f(x) = f(y).
⇒ `x^2 + x +1 = y^2 + y +1`
⇒ `(x^2 - y^2 ) + (x - y ) = 0 `
⇒ (x +y) (x- y ) + (x-y ) = 0
⇒ ( x - y) ( x + y + 1) = 0
⇒ x - y = 0 [ x + y + 1 can not be zero because x and y are natural numbers
⇒ x =y
So, f is one-one.
Surjectivity:
when x = 1
`x^2 + x +1 = 1 +1 +1 = 3`
⇒ x + x +1 ≥ 3 , for every x in N.
⇒ f(x) will not assume the values 1 and 2.
So, F is not onto.
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