About Exercise 2.3.1: Proving conditional statements by contrapositive Prove each statement by contrapositive

a. For every integer n, if n2 is an odd, then n is odd. Solution

b. For every integer n, if n is even, then n is even

c. For every integer n, if 5n + 3 is even, then n is odd.

d. For every integer n, if n2 2n 7 is even, then n is odd

e. For every real number r, if is irrational, then - is also irrational.

f. For every non-zero real number az, if z is irrational, then 1 is also irrational.

g. For every pair of real numbers x and y, lf Z3 + Xy2-ry + y3, then z < y

h. For every integer n, if n2 is not divisible by 4, then n is odd.

i. For every pair of real numbers z and y. if + y is irrational, then z is irrational or y is irrational.

See proofs below

Step-by-step explanation:

A proof by counterpositive consists on assuming the negation of the conclusion and proving the negation of the hypothesis.

a) Assume that n is not odd. Then n is even, that is, n=2k for some integer k. Hence n²=4k²=2 (2k²) = 2t for some integer t=2k². Then n² is even, therefore n² is not odd. We have proved the counterpositive of this statement.

b) Assume that n is not even, then n is odd. Thus, n=2k+1 for some integer k. Now, n³ = (2k+1) ³=8k³+6k²+6k+1=2 (4k³+3k²+3k) + 1=2t+1 for the integer t=4k³+3k²+3k. Thus n³ is odd, that is, n³ is not even.

c) Suppose that n is not odd, that is, n is even. Now, n=2k for some integer k. Then 5n+3=10k+3=2 (5k+1) + 1, thus 5n+3 is odd, then 5n+3 is not even.

d) Suppose that n is not odd, then n is even. Now, n=2k for some integer k. Then n²-2n+7=4k²-4k+7=2 (2k²-2k+3) + 1. Hence n²-2n+7 is odd, that is, n²-2n+7 is not even.

e) Assume that - r is not irrational, then - r is rational. Since - 1 is rational, then (-1) (-r) = r is rational. Thus r is not irrational.

f) Assume that 1/z is not irrational. Then 1/z is rational. Multiplucative inverses of rational numbers are rational, hence z is rational, that is, z is not irrational.

g) Suppose that z>y. We will prove that z³+zy²≤z²y+y³ is false, that is, we will prove that z³+zy²>z²y+y³. Multiply by the nonnegative number z² in the inequality z>y to get z³>z²y (here we assume z and y nonzero, in this case either z³>0=y³ is true or z³=0>y³ is true). On the other hand, multiply by z² (positive number) to get zy²>y³. Add both inequalities to obtain z³+zy²>z²y+y³ as required.

h) Suppose than n is even. Then n=2k, and n²=4k² is divisible by 4.

i) Assume that "z is irrational or y is irrational" is false. Then z is rational and y is rational. Rational numbers are closed under sum, then z+y is rational, that is, z+y is not irrational.