How can i find a bijection from N( natural numbers) to Q[X]

In summary, the proof is as follows:1. Let A_0 be the set of all natural numbers.2. Let A_1 be the set of all numbers that are the sum of two rational numbers.3. Let A_2 be the set of all numbers that are the sum of two rational numbers and the square of a rational number.4. Repeat the process for each larger set of numbers.5. Finally, combine all the sets together to get Q[x]. This is a countable set, and thus there is a bijection from N to Q[x].
  • #1
cghost
4
0
How can i find a bijection from N( natural numbers) to Q[X] ( polynomials with coefficient in rational numbers ). I can't find a solution for this. Can you please point me in the right direction ?
 
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  • #2


This is a standard trick. The proof is as follows:

Let [tex]A_0=\mathbb{Q}[/tex]. This is clearly countable.
Let [tex]A_1=\{a+bx~\vert~a,b\in \mathbb{Q}\}[/tex], this is countable since it has the same cardinality of [tex]\mathbb{Q}\times \mathbb{Q}[/tex].
Let [tex]A_2=\{a+bx+cx^2~\vert~a,b,c\in \mathbb{Q}\}[/tex], this is countable since it has the same cardinality of [tex]\mathbb{Q}\times\mathbb{Q}\times\mathbb{Q}[/tex].

Continue with this process. Finally, we have [tex]\mathbb{Q}[x]=\bigcup_{n\in \mathbb{N}}{A_n}[/tex]. This is countable as countable union of countable sets.
 
  • #3


micromass said:
This is a standard trick. The proof is as follows:

Let [tex]A_0=\mathbb{Q}[/tex]. This is clearly countable.
Let [tex]A_1=\{a+bx~\vert~a,b\in \mathbb{Q}\}[/tex], this is countable since it has the same cardinality of [tex]\mathbb{Q}\times \mathbb{Q}[/tex].
Let [tex]A_2=\{a+bx+cx^2~\vert~a,b,c\in \mathbb{Q}\}[/tex], this is countable since it has the same cardinality of [tex]\mathbb{Q}\times\mathbb{Q}\times\mathbb{Q}[/tex].

Continue with this process. Finally, we have [tex]\mathbb{Q}[x]=\bigcup_{n\in \mathbb{N}}{A_n}[/tex]. This is countable as countable union of countable sets.

You proved here that there is a bijection from N to Q[x], but how can i write that function down ?
Or you can't exactly explicit the function f : N -> Q[x] , f(x) = ... ?
If it's a bijection, every natural number should point to a single polynom, is that right?
Or Q[X] is defined as a set of equivalence classes.
 
  • #4


It's very hard to explicitely say what the bijection is. Most of the time people are happy with knowing that there exists a bijection.

Of course, if you examine all the steps I made, maybe you can make the bijection step-by-step. But this will be a very ugly thing. In fact, even a bijection between [tex]\mathbb{N}[/tex] and [tex]\mathbb{Q}[/tex] is hard and ugly to write down...
 
  • #5


Thanks for helping me
 

FAQ: How can i find a bijection from N( natural numbers) to Q[X]

1. How can I prove that there exists a bijection between N and Q[X]?

To prove the existence of a bijection between two sets, we need to show that there is a one-to-one and onto mapping between them. In this case, we can use the Cantor-Bernstein-Schroeder theorem, which states that if there exist injective functions from N to Q[X] and from Q[X] to N, then there exists a bijection between the two sets. Therefore, we just need to find such functions and show that they are both injective.

2. What is the function that maps N to Q[X] in a bijection?

The function that maps N to Q[X] in a bijection is f(n) = xn. This function takes a natural number n and maps it to the polynomial xn in Q[X]. It is one-to-one because different natural numbers will always map to different polynomials, and it is onto because every polynomial in Q[X] can be expressed as xn for some natural number n.

3. Can you explain why the set of natural numbers and the set of polynomials in Q[X] have the same cardinality?

Yes, the two sets have the same cardinality because we have just shown that there exists a bijection between them. This means that they have the same number of elements, even though they may seem to be of different sizes at first glance. This is one of the interesting properties of infinite sets.

4. Is the bijection between N and Q[X] unique?

No, the bijection between N and Q[X] is not unique. In fact, there are infinitely many bijections that can be defined between these two sets. For example, we could define a different bijection by mapping each natural number n to the polynomial xn+1. As long as the function is one-to-one and onto, it is a valid bijection.

5. How can understanding bijections between sets help in solving mathematical problems?

Understanding bijections between sets is useful in many areas of mathematics, such as combinatorics, number theory, and topology. It allows us to determine the cardinality of sets and to establish relationships between seemingly different sets. In particular, it can be helpful in proving theorems and solving problems involving infinite sets, where traditional methods may not work. It also provides a deeper understanding of the structure of sets and their elements.

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