Question

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. Suppose m is any even integer and n is any odd integer. If m · n is even, then by definition of even there exists an integer r such that m · n = 2r. By definition of even and odd, there exist integers p and q such that m = 2p and n = 2q + 1. Therefore, by substitution, the product m · n = (2p)(2q + 1) = 2r. But r is an integer by the statement in Step 2. Thus m · n is two times an integer, so by definition of even, the product is even. Identify the mistakes in the proof.

Step 1 begins an argument by example using specific values m and n
Step 2 assumes what needs to be proved about m n at the end.
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 is not actually known to be true.
Step 5 relies on a previous incorrect step that assumes what needs to be proved
Step 6 draws a conclusion about m n that belongs in the hypothesis

Answer 1

Answer:

When specific values were put for m and n, it was consistent with the theorem.

That is, m is even integer and n is odd integer.

The product m . n is even, as shown below.

The equivalent expression m . n = (2p)(2q + 1) = 2r is true.

Step-by-step explanation

To check for mistake(s) in the proof of the theorem, we assume specific values for m, n, p, q and r.

Let us put

m = 4(An even integer)

n = 7(An odd integer)

p = 2, q = 3. p and q are integers

Then m = 2p = 2 x 2 = 4

n = 2q + 1 = (2 x 3) + 1 = 6 + 1 = 7

Therefore,

m . n = (2p)(2q + 1) = 2r

m . n = 4 x 7 = [(2x2)*(2x3 + 1)]

m . n = 28(2*14). Here r is 14