My questions regarding the below theorem
Both questions are centred on Eq(2) and the paragraph preceding it.
1) How is it that Eq(2) contains $a_k$ but in that section of the proof the assumption is that the theorem is true for n≤k-1. Is not this assumption being overreached by the equation running to $a_k$ and therefore invalid?
My own attempt to answer this is that due to the axiom of addition mentioned in the preamble it is always reasonable to add one more element if it is available - but I am unconvinced.
2) How are the contents of the square brackets derived in RHS Eq(2)?
My own attempt to answer: the $a_{q-1}$ reflects the $a_{k-1}$ in the preceding text, and thus $a_k$ is 'broken off' the end of $[(a_1 + ... + a_{q-1})+a_q]$ which is a legitimate process related to axiom I the addition axiom mentioned below. This does not seem very satisfactory.
All help appreciated, thank you.
From An Introduction to the Theory of Groups by Paul Alexandroff
Extracts and paraphrases from the preamble
Axioms of addition include:
I. Two numbers can be added together (i.e. to any two numbers a and b there corresponds a uniquely determined number which we call their sum: a+b.
Later the Assoc Law is defined (a+b)+c = a+(b+c) and then it is established that this defines the sum a+b+c.
[We can conveniently define the sum of four elements a,b,c,d to be equal to a+(b+c+d) i.e.:
$a+b+c+d = a+(b+c+d)\qquad$ (0)
The author then goes on to generalise the result to n elements as follows]
Now we assume that the sum of any $(n-1)$ elements has already been defined; then we define the sum of the n elements $a_1 + ... + a_n$ to be $a_1+(a_2 + ... + a_n)$ as being defined for arbitrary n by the method of complete induction.
Theorem
Let n be any natural number, for every natural number $m≤n$ we have
$(a_1 + ... + a_m)\,+(a_{m+1}+ ... + a_n)=a_1+ ... + a_n\qquad$(1)
Proof. The proof will proceed by the method of complete induction. For n=1 the theorem simply states that $a_1 = a_1$. We assume that it is true for $n≤k-1$, and prove it for n = k. We consider first the case m = 1. Then (1) becomes
$a_1 +(a_2+ ... + a_k)\,=a_1+ ... + a_k$
But this is just the definition of the expression $a_1+ ... + a_k$ [see (0)] Therefore (1) is true for n=k and m=1.
Now let us still consider n=k and let us assume that our formula (1) is proved for m=q-1; we prove it for m=q. Since (1) is obviously true for m=n, we can assume that $q<k$. Since the truth of the theorem for $n≤k-1$ is assumed, we have
$(a_1 + ... + a_q)+(a_{q+1}+ ... + a_k)= [(a_1 + ... + a_{q-1})+a_q]+(a_{q+1}+ ... + a_k)\qquad$(2)
The associative law, applied to the three elements $(a_1 + ... + a_{q-1}),(a_q),(a_{q+1} + ... + a_k)$ gives
$[(a_1 + ... + a_{q-1})+a_q]+(a_{q+1} +... + a_k)= (a_1 + ... + a_{q-1})+[a_q+(a_{q+1}+ ... + a_k)]$
But the expression in square brackets on the RHS is by definition equal to
$a_q+a_{q+1}+...+a_k$
Therefore we have
$(a_1 + ... + a_q)+(a_{q+1} +... + a_k)= (a_1 + ... + a_{q-1})+(a_q + ... + a_k)$
[This completes the inductive proof to validate any value of $m$ from $1$ to $k-1$ to split the expression - this is not in the published proof but due to a conversation with Joffan below].
But since the formula (1) is assumed to hold for $n=k\, and\, m=q-1$, the RHS of this last equation is equal to $a_1 +...+ a_k\qquad$ Therefore
$(a_1 + ... + a_m)\,+(a_{m+1}+ ... + a_k)=\,a_1+ ... + a_k$
which completes the induction and gives
$(a_1 + ... + a_m)\,+(a_{m+1}+ ... + a_n)=\,a_1+ ... + a_n$
Which is what we set out to prove.
In answer to your specific points,
1) The inductive assumption is not being applied to the whole expression, but only to the first bracket, which runs to some smaller number than $k$. At that point, the intent is to show that the final element from the first bracket can be shifted into the start of the second bracket.
2) As above, the square bracket represents the action of the inductive assumption on the first bracket, which says we can split the sum at any point - here, we're splitting it one element from the end.
At this point in the proof, we haven't shown that these two expression are equal to the defined sum - that doesn't happen until we repeatedly apply this to leave only one element at the start, which is the definition of the unbracketed expression.