NO NO NO
In the second step, we are NOT trying to prove that k is a member of L. In the first step, we proved 1 is a member of L, so L is not the empty set. There is at least one member of L, namely 1, and we HOPE to prove that there are an infinite number of members, but at this point we do not know if there is one, more than one but less than an infinite number, an infinite number but not all the natural numbers, or what. We just know that one or more numbers are members of L. We choose an ARBITRARY one of those numbers in L and call it k. Because it is arbitrary we cannot ascribe any numeric value to it. But because it is in L we KNOW that it is a natural number and that it has all the antecedent properties PLUS property Q. Furthermore we know that (k + 1) is a natural number and has all the antecedent properties, but we do NOT know whether it has property Q. That is what we must prove.
Let's go back to the example I started in my previous post and do the second step.
Suppose we want to prove the following property of the natural numbers:
\(\displaystyle For\ every\ natural\ number\ x, (x + 4)^2 < 2^{(x + 4)}.\)
We started with:
\(\displaystyle (A)\ Let\ L = the\ set\ of\ all\ natural\ numbers\ x\ such\ that\ (x + 4)^2 < 2^{(x + 4)}\)
We ended the first step with:
\(\displaystyle (C)\ 1 \in L.\)
On to the second step
\(\displaystyle (D) Let\ k\ be\ an\ arbitrary\ member\ of\ L.\) We know that there must be at least one member of L from step 1.
\(\displaystyle (E)\ k\ is\ a\ natural\ number \ge 1.\)
The line above follows from lines D and A and the antecedent property that all natural numbers are greater than or equal to 1 (according to the the more usual definition of the natural numbers).
\(\displaystyle k^2 \ge 1^2 = 1.\) Antecedent property of natural numbers.
\(\displaystyle 8k > 2k.\) Antecedent property of natural numbers.
\(\displaystyle 16 > 9.\) Antecedent property of natural numbers.
\(\displaystyle So\ (k + 4)^2 = k^2 + 8k + 16 \ge 1 + 8k + 16 > 8k + 16 > 2k + 9.\)
\(\displaystyle Or\ (2k + 9) < (k + 4)^2.\)
\(\displaystyle So\ (k + 4)^2 + (2k + 9) < (k + 4)^2 + (k + 4)^2.\)
\(\displaystyle Or\ k^2 + 8k + 16 + 2k + 9 < 2(k + 4)^2.\)
\(\displaystyle Or\ k^2 + 10k + 25 < 2(k + 4)^2.\)
\(\displaystyle \ Or\ (k + 5)^2 < 2(k + 4)^2.\)
\(\displaystyle (F)\ Or\ [(k + 1) + 4]^2 < 2(k + 4)^2.\)
\(\displaystyle (k + 4)^2 < 2^{(k + 4)}.\) This follows from lines D and A. Notice that k has been substituted for x in the definition of property Q.
\(\displaystyle So\ 2(k + 4)^2 < 2(2^{(k + 4)}).\) Antecedent property.
\(\displaystyle Or\ 2(k + 4)^2 < 2^{(k + 5)}.\) Antecedent property.
\(\displaystyle (G)\ Or\ 2(k + 4)^2 < 2^{[(k + 1)+4]}.\) Antecedent property.
\(\displaystyle (H)\ So\ [(k + 1) + 4]^2 < 2^{[(k + 1) + 4]}.\)
Line H follows from lines F and G. It is the critical part of the proof. We have just shown that (k + 1) has property Q. Notice that (k + 1) has been substituted for x in the definition of property Q.
\(\displaystyle (J)\ And\ (k + 1)\ is\ a\ natural\ number.\)
The line above represents an antecedent property of k, which is a natural number by line E: the succesor of a natural number is also a natural number
\(\displaystyle (R) So\ (k + 1) \in L.\) This follows from lines H, J, and A.
\(\displaystyle So\ L = the\ set\ of\ all\ natural\ numbers.\) This follows from lines C, D, and R.
\(\displaystyle THUS,\ for\ any\ natural\ number\ x, (x + 4)^2 < 2^{(x+4)}.\)
I'll bet that was not an example your teacher gave.
