Asked by Samantha
Let A= {for all m that's an element of the integers | m=3k+7 for some k that's an element of positive integers}. Prove that A is countably infiite. Note: you must define a function from Z+ to A, and then prove that the function you definied is a bijection
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MathMate
So the question is:
A={m∈ℤ : m=3k+7 ∃k∈ℤ<sup>+</sup>}
Prove that A is countably infinite by defining a function f : ℤ -> A and proving that f is bijective.
A function f is bijective if and only if its inverse is also a function.
Since the inverse of f is relative easy to find, all you need is to prove that the inverse is a function.
If f is bijective (i.e. one-to-one and onto), and if the domain is countably infinite, the range of f, i.e. A, must be also.
A={m∈ℤ : m=3k+7 ∃k∈ℤ<sup>+</sup>}
Prove that A is countably infinite by defining a function f : ℤ -> A and proving that f is bijective.
A function f is bijective if and only if its inverse is also a function.
Since the inverse of f is relative easy to find, all you need is to prove that the inverse is a function.
If f is bijective (i.e. one-to-one and onto), and if the domain is countably infinite, the range of f, i.e. A, must be also.
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