For any n, m given ##I_n \thicksim I_m##, prove that ## n=m##

[issacnewton]

Homework Statement

Prove that for all natural numbers n and m, if $I_n \thicksim I_m$, then $n=m$.

Relevant Equations

Definition of equinumerous sets and definition of bijection

Let $n, m$ be arbitrary in $\mathbb{N}$. Assume that $I_n \thicksim I_m$. We are given that

$$ I_n = \big\{ i \in \mathbb{Z}^+ | i \leq n \big\} $$
$$ I_m = \big\{ i \in \mathbb{Z}^+ | i \leq m \big\} $$

These are following sets

$$ I_n = \bigg \{ 1,2,3,\ldots, n \bigg \} $$
$$ I_m = \bigg \{ 1,2,3,\ldots, m \bigg \} $$

Now, assume that $n < m$. Since $I_n \thicksim I_m$, there is a bijection from $I_n$ to $I_m$. Since there are less members in $I_n$ than in $I_m$, we can map these $n$ members to first $n$ members in $I_m$. Then we are left with $m-n$ members in $I_m$ which do not have any pre image in $I_n$. But since the function is onto, this is a contradiction and hence $n \nless m$. Similar argument can be given when $n > m$. So we also prove that $n \ngtr m$. So, we must have $n = m$. Is this a valid proof ?

 

[Mark44]

Homework Statement:: Prove that for all natural numbers n and m, if $I_n \thicksim I_m$, then $n=m$.
Relevant Equations:: Definition of equinumerous sets and definition of bijection

Let $n, m$ be arbitrary in $\mathbb{N}$. Assume that $I_n \thicksim I_m$. We are given that

$$ I_n = \big\{ i \in \mathbb{Z}^+ | i \leq n \big\} $$
$$ I_m = \big\{ i \in \mathbb{Z}^+ | i \leq m \big\} $$

These are following sets

$$ I_n = \bigg \{ 1,2,3,\ldots, n \bigg \} $$
$$ I_m = \bigg \{ 1,2,3,\ldots, m \bigg \} $$

Now, assume that $n < m$. Since $I_n \thicksim I_m$, there is a bijection from $I_n$ to $I_m$. Since there are less members in $I_n$ than in $I_m$, we can map these $n$ members to first $n$ members in $I_m$. Then we are left with $m-n$ members in $I_m$ which do not have any pre image in $I_n$. But since the function is onto, this is a contradiction and hence $n \nless m$. Similar argument can be given when $n > m$. So we also prove that $n \ngtr m$. So, we must have $n = m$. Is this a valid proof ?

Looks fine to me.

 

[issacnewton]

Thanks Mark