- #1

sleepingMantis

- 25

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<Moderator's note: Moved from a technical forum.>

Hi PF,

I am learning how to prove things (I have minimal background in math). Would the following proof be considered valid and rigorous? If not any pointers or tips would be much appreciated!

Prove that the notion of number of elements of a nonempty finite set is a well defined concept. More precisely, prove that there exists a bijection ## f : I_m → I_n## if and only if ##m = n##.

## I_n ## denotes the set: ## I_n = \{ j \in \mathbb{N} ; 1 \leq j \leq n \} ##

First prove that ## f : I_m → I_n \Rightarrow m = n##:

Assume that ## f ## is a bijection from the set ## I_m ## to the set ## I_n ##, so that we can write ## f : I_m → I_n##

Writing ## I_n = \{ a_1, ..., a_n \} ## and similarly ## I_m = \{ b_1, ..., b_m \} ## for some ##m, n \in \mathbb{N} ##, since ## f ## is a bijection, for every element ## a_i \in I_n ## and every element ##b_k \in I_m## we have ## f(a_i) = b_k ## for some ##i, k##. Since ## f ## is one to one, it follows that ## |I_n| = |I_m| ##. Therefore ##m = n##.

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Now prove that ##n = m \Rightarrow f : I_m → I_n##:

Assume ## m = n ##.

Construct two sets, ##I_n = \{ a_1, ..., a_n \} ## and ## I_m = \{ b_1, ..., b_m \} ##. Since we have the same number of elements, we can create a bijection ##f: I_n → I_m ##.

so ##n = m \Rightarrow f : I_m → I_n##

Hi PF,

I am learning how to prove things (I have minimal background in math). Would the following proof be considered valid and rigorous? If not any pointers or tips would be much appreciated!

**Problem:**Prove that the notion of number of elements of a nonempty finite set is a well defined concept. More precisely, prove that there exists a bijection ## f : I_m → I_n## if and only if ##m = n##.

**Relevant Definitions:**## I_n ## denotes the set: ## I_n = \{ j \in \mathbb{N} ; 1 \leq j \leq n \} ##

**Attempt at proof:**First prove that ## f : I_m → I_n \Rightarrow m = n##:

Assume that ## f ## is a bijection from the set ## I_m ## to the set ## I_n ##, so that we can write ## f : I_m → I_n##

Writing ## I_n = \{ a_1, ..., a_n \} ## and similarly ## I_m = \{ b_1, ..., b_m \} ## for some ##m, n \in \mathbb{N} ##, since ## f ## is a bijection, for every element ## a_i \in I_n ## and every element ##b_k \in I_m## we have ## f(a_i) = b_k ## for some ##i, k##. Since ## f ## is one to one, it follows that ## |I_n| = |I_m| ##. Therefore ##m = n##.

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Now prove that ##n = m \Rightarrow f : I_m → I_n##:

Assume ## m = n ##.

Construct two sets, ##I_n = \{ a_1, ..., a_n \} ## and ## I_m = \{ b_1, ..., b_m \} ##. Since we have the same number of elements, we can create a bijection ##f: I_n → I_m ##.

so ##n = m \Rightarrow f : I_m → I_n##

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