Let f:ℤ+ → ℤ+ be the function defined by: for each x ∈ ℤ+, f(x) is the number of positive divisors of x.
- find integers m, n, nd k so tha f(m)=3, f(n) = 4 anf f(k) = 5
-is f one-to one? Explain, is f unto? prove
is it true that all positive odd integers a and b, f(ab) = f(a) f(b)? prove answer
To find integers m, n, and k that satisfy the conditions:
1. f(m) = 3:
Start by listing numbers with 3 positive divisors: 4, 9, 25, ...
Choose any number from the list, for example, m = 4.
2. f(n) = 4:
List numbers with 4 positive divisors: 6, 8, 10, ...
Choose any number from the list, for example, n = 6.
3. f(k) = 5:
List numbers with 5 positive divisors: 16, 18, 20, ...
Choose any number from the list, for example, k = 16.
Now we have m = 4, n = 6, and k = 16.
To determine if f is a one-to-one (injective) function, we need to check if different inputs map to different outputs. In other words, if f(a) = f(b), then a = b. To show it is one-to-one, we need to prove this statement.
To prove one-to-one:
Let's assume f(a) = f(b) for some positive integers a and b.
That means the numbers of divisors for a and b are the same.
If a and b have the same number of divisors, it implies that a and b are either equal or have the same prime factorization.
Considering a = p₁^a₁ * p₂^a₂ * ... * pn^an and b = q₁^b₁ * q₂^b₂ * ... * qm^bm:
- If a ≠ b, then a and b should have different prime factors, which means they cannot have the same number of divisors.
- Therefore, f(a) = f(b) implies a = b.
Since f(a) = f(b) implies a = b, the function f is one-to-one (injective).
To determine if f is onto (surjective), we need to check if every possible output has at least one corresponding input. In other words, for any y ∈ ℤ+, there exists an x ∈ ℤ+ such that f(x) = y.
To prove onto:
Let's consider any positive integer y.
To find the corresponding x, we need a number with y positive divisors.
One way to construct such a number is to take a prime number to the power of y-1, i.e., x = p^(y-1).
For example, if y = 4, then x = 2^3 = 8. This means f(8) = 4, which satisfies the condition.
Since we can always find an x for any positive integer y, f is onto (surjective).
Now let's determine if f(ab) = f(a) * f(b) for all positive odd integers a and b.
To prove if f(ab) = f(a) * f(b):
Let's consider an odd number a and an odd number b.
Prime factorize a as a = p₁^a₁ * p₂^a₂ * ... * pn^an.
Prime factorize b as b = q₁^b₁ * q₂^b₂ * ... * qm^bm.
When we multiply a and b, the resulting number can be represented as ab = p₁^a₁ * p₂^a₂ * ... * pn^an * q₁^b₁ * q₂^b₂ * ... * qm^bm.
The number of divisors of ab will be the product of the powers of each prime factor plus 1:
f(ab) = (a₁ + 1)*(a₂ + 1)*...*(an + 1)*(b₁ + 1)*(b₂ + 1)*...*(bn + 1).
Similarly, the number of divisors of a will be:
f(a) = (a₁ + 1)*(a₂ + 1)*...*(an + 1).
And the number of divisors of b will be:
f(b) = (b₁ + 1)*(b₂ + 1)*...*(bn + 1).
Therefore, f(ab) = f(a) * f(b).
This proves that for all positive odd integers a and b, f(ab) = f(a) * f(b).
Note: The proof provided above is a logical explanation of why the property holds true for all positive odd integers a and b.