Question

not
if A f: (zxz)-->z
is defined by
f(x) = 4x+5y Prove that f i
s
one to one but onto .​

Answer 1

Answer:

Proof. First, let’s check that f is injective. Suppose f (x) = f (y). Then

5x+1 5y+1

x−2 = y−2 (5x+1)(y−2) = (5y+1)(x−2)

5xy−10x+y−2 = 5yx−10y+x−2 −10x+y = −10y+x

11y = 11x

y = x.

Explanation:

Since f (x) = f (y) implies x = y, it follows that f is injective.

Next, let’s check that f is surjective. For this, take an arbitrary element b ∈ R−{5}.

Wewanttoseeifthereisanx∈R−{2}forwhich f(x)=b,or 5x+1 =b. Solvingthis x−2

for x, we get:

5x+1 = b(x−2)

5x+1 = bx−2b 5x−xb = −2b−1 x(5−b) = −2b−1.