Homepage Solution manuals Ivan Niven An Introduction to the Theory of Numbers Exercise 7.1.5 ($\langle a_0,a_1,\ldots,a_n \rangle > \langle a_0,a_1,\ldots,a_n + c \rangle$ if $n$ odd.)

Exercise 7.1.5 ($\langle a_0,a_1,\ldots,a_n \rangle > \langle a_0,a_1,\ldots,a_n + c \rangle$ if $n$ odd.)

Let a 1 , a 2 , , a n and c be positive real numbers. Prove that

a 0 , a 1 , , a n > a 0 , a 1 , , a n + c

holds if n is odd, but is false if n is even.

Answers

Proof. We prove this property by induction.

If n = 0 , then a 0 = a 0 < a 0 + c = a 0 + c .

If n = 1 , then a 0 , a 1 > a 0 , a 1 + c by Problem 4.

We define 𝒫 ( k ) by

𝒫 ( k ) c , ( a 0 , a 1 , , a 2 k + 1 ) ( ) 2 k + 2 , { a 0 , a 1 , a 2 k < a 0 , a 1 , , a 2 k + c , a 0 , a 1 , , a 2 k + 1 > a 0 , a 1 , , a 2 k + c .

We have verified that 𝒫 ( 0 ) is true. Suppose now that 𝒫 ( k ) is true, and let a 0 , a 1 , , a 2 k + 2 , a 2 k + 3 be positive real numbers. Applying 𝒫 ( k ) to the ( 2 k + 2 ) -uple ( a 1 , a 2 , , a 2 k + 2 ) , we know that for any c > 0 ,

a 1 , a 2 , , a 2 k + 2 > a 1 , a 2 , , a 2 k + 2 + c > 0 .

Therefore

a 0 , a 1 , , a 2 k + 2 = a 0 + 1 a 1 , , a 2 k + 2 < a 0 + 1 a 1 , , a 2 k + 2 + c = a 0 , a 1 , , a 2 k + 2 ,

so

a 0 , a 1 , , a 2 k + 2 < a 0 , a 1 , , a 2 k + 2 + c . (1)

Applying (1) to the ( 2 k + 3 ) -uple ( a 1 , a 2 , , a 2 k + 3 ) , we know that

0 < a 1 , a 2 , , a 2 k + 3 < a 1 , a 2 , , a 2 k + 3 + c .

Therefore

a 0 , a 1 , , a 2 k + 3 = a 0 + 1 a 1 , , a 2 k + 3 > a 0 + 1 a 1 , , a 2 k + 3 + c = a 0 , a 1 , , a 2 k + 3 ,

so

a 0 , a 1 , , a 2 k + 3 > a 0 , a 1 , , a 2 k + 3 . (2)

By (1) and (2), we obtain that 𝒫 ( k + 1 ) is true, and the induction is done.

This shows that if a 0 , a 1 , , a n and c are positive real numbers, then

a 0 , a 1 , , a n > a 0 , a 1 , , a n + c  if  n  is odd, a 0 , a 1 , , a n < a 0 , a 1 , , a n + c  if  n  is even.

Unfortunately, the statement don’t assume that a 0 > 0 , so we apply again the same argument: if a is any real number, and a 1 , , a n positive real numbers, since

a 1 , , a n > a 1 , , a n + c  if  n 1  is odd, a 1 , , a n < a 1 , , a n + c  if  n 1  is even.

then

a 0 + 1 a 1 , , a n < a 0 + 1 a 1 , , a n + c  if  n 1  is odd, a 0 + 1 a 1 , , a n > a 0 + 1 a 0 , a 1 , , a n + c  if  n 1  is even,

so

a 0 , a 1 , , a n < a 0 , a 1 , , a n + c  if  n  is even , a 0 , a 1 , , a n > a 0 , a 1 , , a n + c  if  n  is odd.

(even if a 0 < = 0 ). □

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2025-07-23 10:27
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