Question

Prove that these four statements about the integer n are equivalent:

(i) \({n^2}\)is odd,

(ii)\(1 - n\) is even,

(iii) \({n^3}\)is odd,

(iv) \({n^2} + 1\)is even.

Short answer

The four given statements (i), (ii), (iii) and (iv) are equivalent.

Step-by-step solution

Introduction

Integral values are even if they are in the form of\(2n\)and that are odd if they are in the form of \(2n + 1\)

To prove (i) is equivalent to (ii)

Suppose that \({n^2}\)is odd

\( \Rightarrow n.n\)is odd

\( \Rightarrow n\)is odd. (since the multiplication of two odd integers is odd)

\( \Rightarrow n = 2k + 1\), for some integer\(k\)

\( \Rightarrow 1 - n = - 2k\)

\( \Rightarrow 1 - n = 2\left( { - k} \right)\) (\(k\)is integer\( \Rightarrow - k\)is also an odd integer)

As \(1 - n\)is even, i.e. statement (i) implies statement (ii).

To prove (ii) is equivalent to (iii)

Suppose \(1 - n\) is even

By the definition of even numbers,

\( \Rightarrow 1 - n = 2k\), for some integral value \(k\)

\( \Rightarrow n = - 2k + 1\)

\( \Rightarrow n = 2\left( { - k} \right) + 1\) (-k is an integral value)

\( \Rightarrow n\)is odd. (since the odd integer is of the form\(2k + 1\) )

\( \Rightarrow n.n\)is odd (since the multiplication of two odd integers is odd)

\( \Rightarrow {n^2}\)is odd

\( \Rightarrow {n^2}.n\) is odd (since the multiplication of two odd integers is odd)

\( \Rightarrow {n^3}\)is odd

Hence, statement (ii) implies statement (iii)

To prove (iii) is equivalent to (iv)

Suppose\({n^3}\)is odd

\( \Rightarrow {n^2}.n\) is odd (since\({n^3} = {n^2}.n\))

\( \Rightarrow {n^2}\)is odd (since the multiplication of two odd integers is odd)

\( \Rightarrow {n^2} = 2k + 1\), for some integer \(k\)

\( \Rightarrow {n^2} + 1 = 2k + 1 + 1\) (add 1 on both sides)

\( \Rightarrow {n^2} + 1 = 2k + 2\)

\( \Rightarrow {n^2} + 1 = 2\left( {k + 1} \right)\), is even

\( \Rightarrow {n^2} + 1\) is even

Hence, statement (iii) implies to statement (iv)

To prove statement (iv) is equivalent to statement (i)

Suppose \({n^2} + 1\)is even.

By the definition of even numbers,

\( \Rightarrow {n^2} + 1 = 2k\), for some integer \(k\).

\( \Rightarrow {n^2} = 2k - 1\), is odd (since odd number is of form\(2k - 1\))

Therefore\({n^2}\)is odd .

Hence, statement (iv) implies to statement (i).

Thus, the given four statements (i), (ii), (iii) and (iv) are equivalent.