Question -
Answer -
Givenf:┬аN┬атЖТ┬аN, defined by┬аf(x) =┬аx2┬а+┬аx┬а+1
Now we have to provethat given function is one-one
Injectivity:
Let┬аx┬аand┬аy┬аbe any two elements in the domain (N), suchthat┬аf(x) = f(y).┬а
тЗТ x2┬а+x + 1 = y2┬а+ y + 1
тЗТ (x2┬атАУy2) + (x тАУ y) = 0 `
тЗТ (x + y) (x- y ) + (xтАУ y ) = 0
тЗТ (x тАУ y) (x + y + 1) =0┬а
тЗТ x тАУ y = 0 [x + y +1┬аcannot be zero because x and y are natural numbers┬а
тЗТ x = y
So,┬аf┬аisone-one.
Surjectivity:┬а
When x = 1
x2┬а+ x+ 1 = 1 + 1 + 1 = 3
тЗТ x2┬а+x +1 тЙе 3,┬аfor every x in N.
тЗТ┬аf(x) will notassume the values 1 and 2.┬а
So, f is not onto.