let f: ℤ→ ℤ and g: ℤ → ℤ be functions defined by f(x) = 3x + 1 and g(x) = |_x/2_|.
(a) Is f o g one-to-one? prove
(b) is f o g unto? prove
(c) Is g o f one-to-one?
(d) Is g o f unto?
Thank you for your help.
clearly g is not 1-1. For example, g(3) = g(2)
Also, it is clear that both f and g are onto.
Some good discussion can be found here:
http://www.csee.umbc.edu/~stephens/203/PDF/7-3.pdf
f◦g = 3g+1 = 3⌊x/2⌋+1
g◦f = ⌊f/2⌋ = ⌊(3x+1)/2⌋
g◦f is clearly 1-1, since consecutive values of x produce 3x+1 which are separated by more than 2, making their floors distinct.
Oops. g◦f and f◦g are not onto, since
(g◦f)(0) = 0
(g◦f)(1) = 2
And there is no x such that (g◦f)(x) = 1
(f◦g)(0) = 1
(f◦g)(1) = 1
(f◦g)(2) = 4
and there is no (f◦g)(x) = 2
To determine whether a composite function is one-to-one or onto, we need to consider its individual components and their properties. Let's break down each part and analyze them.
(a) Is f o g one-to-one? To prove the one-to-one property, we need to show that for any two distinct inputs, their outputs will also be distinct.
To find f o g, we substitute the function g(x) = |_x/2_| into f(x), resulting in f(g(x)) = f(|_x/2_|).
To prove one-to-one, we assume two different inputs, say a and b, such that a ≠ b. We need to show that f(g(a)) ≠ f(g(b)).
Let's consider two cases:
1. If a and b have different signs (one positive, one negative), then |_a/2_| ≠ |_b/2_|. Since f(x) = 3x + 1, f(g(a)) ≠ f(g(b)) since 3 * |_a/2_| + 1 ≠ 3 * |_b/2_| + 1.
2. If both a and b are either positive or negative, their floors are equal. Let k be the common floor value. In this case, f(g(a)) = f(k) and f(g(b)) = f(k). Since k is the same for both inputs, f(g(a)) = f(g(b)).
Thus, f o g is not one-to-one since there exist inputs a ≠ b for which f(g(a)) = f(g(b)).
(b) Is f o g onto? To prove onto, we need to show that for any output value, there exists at least one input value that maps to it.
Let's consider any arbitrary integer y and find an input value x (or values) that maps to it. To do this, we need to solve the equation y = f(g(x)).
From the definition of g(x) = |_x/2_|, we have |_x/2_| = (y - 1)/3 (after substituting into f(x) = 3x + 1).
The expression |_x/2_| represents the nearest integer less than or equal to x/2. So, |_x/2_| = (y - 1)/3 implies (y - 1)/3 ≤ x/2 < (y - 1)/3 + 1.
By manipulating the inequality, we find that (2y - 4)/3 < x ≤ (2y + 1)/3.
Since x is an integer, we can conclude that there is at least one value of x that maps to y for any given y. Therefore, f o g is onto.
(c) Is g o f one-to-one? To analyze this, we find the composition g o f by substituting f(x) = 3x + 1 into g(x), resulting in g(f(x)) = |_(3x + 1)/2_|.
To prove one-to-one, we assume two different inputs, say p and q, such that p ≠ q. We need to show that g(f(p)) ≠ g(f(q)).
Here we encounter a challenge since the expression |_(3x + 1)/2_| contains the floor function, making it difficult to compare directly. To determine one-to-one, we need to find a counterexample where g(f(p)) = g(f(q)).
Let's consider the specific case: p = 0 and q = -1, which yields f(p) = 1 and f(q) = -2. Now, g(f(p)) = |_(3*0 + 1)/2_| = |_1/2_| = 0, and g(f(q)) = |_(3*-1 + 1)/2_| = |_(-2)/2_| = 0.
Since g(f(p)) = g(f(q)), we have found a counterexample where p ≠ q but g(f(p)) = g(f(q)). Thus, g o f is not one-to-one.
(d) Is g o f onto? To prove onto, we need to show that for any output value, there exists at least one input value that maps to it.
To analyze this, we consider that g o f is a composition of functions, and g(f(x)) = |_(3x + 1)/2_|.
Using similar reasoning as in part (b), we pick any integer y and determine its pre-image by solving the equation g(f(x)) = y.
As before, |_x/2_| = (y - 1)/3 (after substituting f(x) = 3x + 1).
Through the same steps as in part (b), we find (2y - 4)/3 < x ≤ (2y + 1)/3.
Thus, there exists at least one value of x that maps to any given y. Therefore, g o f is onto.
In summary:
(a) f o g is not one-to-one.
(b) f o g is onto.
(c) g o f is not one-to-one.
(d) g o f is onto.