Prove that the function f: NN, defined by f(x) = x2 + x+1 is one one but not onto.

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 f(x) = x2+x+1

 

Let x1, y1 N such that f(x1) = f(y1
 
x12+ x1+1 =y12 + y1 + 1
=(x1 - y1) (x1+y1+1) = 0 [As x1+y1+1  for any N] 
 
x1=y1 f is one-one function
 
Clearly f(x) = x12 + x+1 > 3 for x E N 
 
But f(x) does not assume values 1 and 2 
 
f:N N is not onto function
 
 
 
 
 
f:N N is not onto functionf(x) = x2+x+1
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