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Construct a know-show table for each of the following statements and then write a formal proof for one of the statements.

(a) If m is an odd integer, then m + 1 is an even integer.
(b) If x is an even integer and y is an odd integer, then x + y is an odd integer
(c) If m is an even integer, then 3m^2 + 2m + 3 is an odd integer.

User Cmc
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1 Answer

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Explanation:

(a)

If m is an odd integer, then m + 1 is an even integer.

m (odd integer) m + 1 (even integer)

1 2

3 4

5 6

... ...

Proof:

Suppose m is an odd integer. We can write m as 2n + 1 for some integer n. Then,

m + 1 = (2n + 1) + 1 = 2n + 2

Since 2n + 2 is clearly an even integer, it follows that m + 1 is an even integer if m is an odd integer.

(b)

If x is an even integer and y is an odd integer, then x + y is an odd integer

x (even integer) y (odd integer) x + y (odd integer)

0 1 1

2 3 5

4 5 9

... ... ...

Proof:

Suppose x is an even integer and y is an odd integer. We can write x as 2n and y as 2m + 1 for some integers n and m. Then,

x + y = 2n + (2m + 1) = 2(n + m) + 1

Since n + m is clearly an integer, it follows that x + y is an odd integer if x is an even integer and y is an odd integer.

(c)

If m is an even integer, then 3m^2 + 2m + 3 is an odd integer.

m (even integer) 3m^2 + 2m + 3 (odd integer)

0 3

2 27

4 99

... ...

Proof:

Suppose m is an even integer. We can write m as 2n for some integer n. Then,

3m^2 + 2m + 3 = 3(2n)^2 + 2(2n) + 3 = 12n^2 + 4n + 3

Since 12n^2 + 4n + 3 is clearly an odd integer, it follows that 3m^2 + 2m + 3 is an odd integer if m is an even integer.

User Abuduba
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7.1k points