[proofplan] Choose finite bases of $M$ and $N$, with cardinalities equal to their ranks. Embed these basis elements into $M \oplus N$ through the two coordinate summands, obtaining the concatenated family $(e_i,0_N)$ and $(0_M,f_j)$. We prove directly that this family spans and is linearly independent, so it is a finite basis of $M \oplus N$. The invariant basis number hypothesis then ensures that the cardinality of this basis is exactly the rank of $M \oplus N$. [/proofplan]

[step:Choose finite bases realizing the two ranks] Let \begin{align*} m := \operatorname{rank}_R(M) \end{align*} and \begin{align*} n := \operatorname{rank}_R(N). \end{align*} Because $M$ and $N$ are finite free, choose a basis $(e_i)_{1 \le i \le m}$ of $M$ and a basis $(f_j)_{1 \le j \le n}$ of $N$. If $m=0$ or $n=0$, the corresponding basis is the empty family; throughout the proof, an empty sum is interpreted as the zero element of the relevant module. [/step]

[step:Construct the candidate basis in the direct sum]For each $1 \le i \le m$, define \begin{align*} u_i := (e_i,0_N) \in M \oplus N. \end{align*} For each $1 \le j \le n$, define \begin{align*} v_j := (0_M,f_j) \in M \oplus N. \end{align*} We claim that the concatenated family \begin{align*} (u_1,\dots,u_m,v_1,\dots,v_n) \end{align*} is a basis of the left $R$-module $M \oplus N$. First prove spanning. Let $(x,y) \in M \oplus N$. Since $(e_i)_{1 \le i \le m}$ is a basis of $M$, there are unique coefficients $a_i \in R$ for $1 \le i \le m$ such that \begin{align*} x = \sum_{i=1}^{m} a_i e_i. \end{align*} Since $(f_j)_{1 \le j \le n}$ is a basis of $N$, there are unique coefficients $b_j \in R$ for $1 \le j \le n$ such that \begin{align*} y = \sum_{j=1}^{n} b_j f_j. \end{align*} Using coordinatewise addition and scalar multiplication in $M \oplus N$, we obtain \begin{align*} (x,y)=\sum_{i=1}^{m} a_i u_i+\sum_{j=1}^{n} b_j v_j. \end{align*} Thus the family spans $M \oplus N$. Now prove [linear independence](/page/Linear%20Independence). Suppose coefficients $a_i \in R$ for $1 \le i \le m$ and $b_j \in R$ for $1 \le j \le n$ satisfy \begin{align*} \sum_{i=1}^{m} a_i u_i+\sum_{j=1}^{n} b_j v_j=(0_M,0_N). \end{align*} Expanding the definitions of $u_i$ and $v_j$ gives \begin{align*} \left(\sum_{i=1}^{m} a_i e_i,\sum_{j=1}^{n} b_j f_j\right)=(0_M,0_N). \end{align*} Equality in the [direct sum](/page/Direct%20Sum) is coordinatewise, so \begin{align*} \sum_{i=1}^{m} a_i e_i=0_M \end{align*} and \begin{align*} \sum_{j=1}^{n} b_j f_j=0_N. \end{align*} Because $(e_i)_{1 \le i \le m}$ and $(f_j)_{1 \le j \le n}$ are linearly independent, all coefficients $a_i$ and $b_j$ are zero. Therefore the concatenated family is linearly independent. Hence $(u_1,\dots,u_m,v_1,\dots,v_n)$ is a basis of $M \oplus N$.[/step]

[guided]The direct sum $M \oplus N$ keeps the two coordinates separate, so the natural candidate basis is obtained by placing the basis of $M$ in the first coordinate and the basis of $N$ in the second coordinate. For each $1 \le i \le m$, define \begin{align*} u_i := (e_i,0_N) \in M \oplus N. \end{align*} For each $1 \le j \le n$, define \begin{align*} v_j := (0_M,f_j) \in M \oplus N. \end{align*} We prove that \begin{align*} (u_1,\dots,u_m,v_1,\dots,v_n) \end{align*} is a basis of $M \oplus N$ by checking the two defining properties: spanning and linear independence. For spanning, take an arbitrary element $(x,y) \in M \oplus N$. The first coordinate $x$ lies in $M$, and $(e_i)_{1 \le i \le m}$ is a basis of $M$, so there exist coefficients $a_i \in R$ for $1 \le i \le m$ such that \begin{align*} x = \sum_{i=1}^{m} a_i e_i. \end{align*} Similarly, the second coordinate $y$ lies in $N$, and $(f_j)_{1 \le j \le n}$ is a basis of $N$, so there exist coefficients $b_j \in R$ for $1 \le j \le n$ such that \begin{align*} y = \sum_{j=1}^{n} b_j f_j. \end{align*} The module structure on $M \oplus N$ is coordinatewise. Therefore \begin{align*} \sum_{i=1}^{m} a_i u_i+\sum_{j=1}^{n} b_j v_j = \left(\sum_{i=1}^{m} a_i e_i,\sum_{j=1}^{n} b_j f_j\right). \end{align*} Substituting the displayed expressions for $x$ and $y$ gives \begin{align*} \sum_{i=1}^{m} a_i u_i+\sum_{j=1}^{n} b_j v_j=(x,y). \end{align*} Since $(x,y)$ was arbitrary, the family spans $M \oplus N$. For linear independence, suppose coefficients $a_i \in R$ for $1 \le i \le m$ and $b_j \in R$ for $1 \le j \le n$ satisfy \begin{align*} \sum_{i=1}^{m} a_i u_i+\sum_{j=1}^{n} b_j v_j=(0_M,0_N). \end{align*} Expanding the definitions of $u_i$ and $v_j$ yields \begin{align*} \left(\sum_{i=1}^{m} a_i e_i,\sum_{j=1}^{n} b_j f_j\right)=(0_M,0_N). \end{align*} Equality in $M \oplus N$ is equality in each coordinate, so \begin{align*} \sum_{i=1}^{m} a_i e_i=0_M \end{align*} and \begin{align*} \sum_{j=1}^{n} b_j f_j=0_N. \end{align*} Now the linear independence of the basis $(e_i)_{1 \le i \le m}$ forces $a_i=0_R$ for every $1 \le i \le m$, and the linear independence of the basis $(f_j)_{1 \le j \le n}$ forces $b_j=0_R$ for every $1 \le j \le n$. Thus every linear relation among the $u_i$ and $v_j$ is the zero relation, so the family is linearly independent. We have proved both spanning and linear independence. Therefore the concatenated family \begin{align*} (u_1,\dots,u_m,v_1,\dots,v_n) \end{align*} is a basis of $M \oplus N$.[/guided]

[step:Count the basis elements and identify the rank] The constructed basis of $M \oplus N$ consists of the $m$ elements $u_1,\dots,u_m$ and the $n$ elements $v_1,\dots,v_n$, so it has $m+n$ elements. Therefore $M \oplus N$ is finite free. Because $R$ has invariant basis number, the rank of a finite free left $R$-module is independent of the chosen finite basis. Hence the basis constructed above gives \begin{align*} \operatorname{rank}_R(M \oplus N)=m+n. \end{align*} Substituting the definitions of $m$ and $n$ gives \begin{align*} \operatorname{rank}_R(M \oplus N)=\operatorname{rank}_R(M)+\operatorname{rank}_R(N). \end{align*} This proves both asserted conclusions. [/step]