## Answers

Therefore g is injective but not surjective.

Therefore f is injective but not surjective.

Lel f:NN be the function f(n)=na. defined by a (2,4), then Il n=2. y = f(n) = 4, le (2,4) et b) (5,23) If n=5 then. y = f(n)= 5? 225 +23 le (5,23) ¢1

с C.1).

If n=f, then. y=11)==1 le (1,1)€ } d (-3,9) If n=-3, then y = f(-3) = (-3)* = 9 (-39) Ef. le

2 @ gi NxIN defined by g(min) = 29.39.

1 For m, = m2, nina., thus (m, n) = (maine). 2m.

= 2m2 and 3ni = 3n2 we have =) 9 (m., n.) = g (m2, 12) g is injective g(min)= 23h Now, every re for KEN St g(min)=k, only prime druison of k are 2 andy Thus other numbere left. are othero no Qre le 5., and mo many nol in the range of g Thus eunjective is not Jo

b t. Non → + defined by 1(n.k)= nk Now for (m., k) = (h2, Kg) le nisha and ki=k2 we have nin = nek, = K2 ->K => [Chooks) = {(uk) injective thus f is

Now let PE Zt, Suppore there existe Some n, and k such that El p=nk plk. then. n= may not be an But for arbitrary ki přk integer.

Te. Il p is not of the power, then n f is not eunjective Thus

3 o Want to find function of such that an o the domain off is infinite eet of is surjective o the codomain of t es infinite eet. of a not injective Let the domain be 7{o} which is infinite and the codamain be N which is also an infinite eet. Let t: 21{o} → N be defined by f(n) = lal the Zlfoy Now for me and ne z\{0}, 1(n) = 1-n1 = n = lnl= fm) thus t is not injective and for every REN, na in- f(n)., ne 71603. thus f is surjective

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