Consider the following theorem.
The product of any even integer and any odd integer is even.
Here is a proof of the theorem with at least one incorrect step.
1. Suppose m is any even integer and n is any odd integer.
2. If m · n is even, then by definition of even there exists an integer r such that m · n = 2r.
3. By definition of even and odd, there exist integers p and q such that m = 2p and n = 2q + 1.
4. Therefore, by substitution, the product m · n = (2p)(2q + 1) = 2r.
5. But r is an integer by the statement in Step 2.
6. Thus m · n is two times an integer, so by definition of even, the product is even.
Identify the mistakes in the proof. (Select all that apply.)
Step 1 begins an argument by example using specific values m and n.
Step 2 states what could be true about m · n.
Step 3 should use only one variable for an integer to write m and n as even and odd.
Step 4 asserts an equivalence of expressions that has not yet been established.
Step 5 draws an invalid conclusion from Step 2.
Step 6 draws a conclusion about m · n that belongs in the hypothesis.