Exercise 1.7

Exercise 7: Fix b > 1 , y > 0 , and prove that there is a unique real x such that b x = y , by completing the following outline.

  • For any positive integer n , b n − 1 ≥ n ( b − 1 ) .
  • Hence b − 1 ≥ n ( b 1 ∕ n − 1 ) .
  • If t > 1 and n > ( b − 1 ) ∕ ( t − 1 ) , then b 1 ∕ n < t .
  • If w is such that b w < y , then b w + ( 1 ∕ n ) < y for sufficiently large n ; to see this, apply part (c) with t = y ⋅ b − w .
  • If w is such that b w < y , then b w − ( 1 ∕ n ) > y for sufficiently large n .
  • Let A be the set of all w such that b w < y , and show that x = sup ⁡ A satisfies b x = y .
  • Prove that this x is unique.

Answers

(a): Using the binomial expansion, for any positive integer n ,

b n = ( 1 + ( b − 1 ) ) n = 1 n + n 1 n − 1 ( b − 1 ) + A

where A is a non-negative sum of terms involving higher powers of ( b − 1 ) . Hence

b n − 1 = n ( b − 1 ) + A ≥ n ( b − 1 ) .

(b): Replace b in case (a) with b 1 ∕ n to get b − 1 ≥ n ( b 1 ∕ n − 1 ) .

(c): From case (b) we have

b − 1 t − 1 ( b 1 ∕ n − 1 ) < n ( b 1 ∕ n − 1 ) ≤ b − 1 b 1 ∕ n − 1 < t − 1 b 1 ∕ n < t

(d): By case (c), if n > ( b − 1 ) ∕ ( y b − w − 1 ) , then b 1 ∕ n < y b − w , or b w + ( 1 ∕ n ) < y .

(e): Applying case (c) to t = y − 1 b w > 1 , if n > ( b − 1 ) ∕ ( y − 1 b w − 1 ) , then b 1 ∕ n < y − 1 b w , or y < b w − ( 1 ∕ n ) .

(f): If b x < y , then from case (d) there is a sufficiently large integer n such that b x + ( 1 ∕ n ) < y , that is, x + ( 1 ∕ n ) ∈ A , contradicting x = sup ⁡ A . And if b x > y , then from case (e) there is a sufficiently large integer n such that b x − ( 1 ∕ n ) > y , so that x − ( 1 ∕ n ) is an upper bound of A , contradicting x = sup ⁡ A .

(g): Suppose there are real numbers x 1 < x 2 such that b x 1 = b x 2 . Then b x 1 b x 2 − x 1 = b x 2 = b x 1 , so that b x 2 − x 1 = 1 . Using the definition of real powers given in exercise 6, this means there are positive integers m , n such that ( b m ) 1 ∕ n ≤ 1 . However, since the positive integeral powers of numbers greater than 1 are also greater than 1, this is impossible.

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2023-08-07 00:00
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Proof of ( a ) . Since b > 1 , and thus b n − 1 ≥ 1 , for n ≠ 1 ,

b > 1 − n b n − 1 − n b n − bn > 1 − n b n − 1 > n ( b − 1 ) .

For n = 1 , b − 1 = b − 1 . Therefore, for n ∈ ℕ , b n − 1 ≥ n ( b − 1 ) . □

Proof of ( b ) . Since ( a ) is true for arbitrary b > 1 , we can substitute b 1 ∕ n for b since b 1 ∕ n > 1 as well. Thus, b − 1 ≥ n ( b 1 ∕ n − 1 ) . □

Proof of ( c ) . From ( b ) and that n > ( b − 1 ) ∕ ( t − 1 ) , we get

b − 1 ≥ n ( b 1 ∕ n − 1 ) b − 1 > b − 1 t − 1 ( b 1 ∕ n − 1 ) 1 > b 1 ∕ n − 1 t − 1 t > b 1 ∕ n . □

Proof of ( d ) . Since b w < y , y b − w > 1 , and let t = y b − w . Then apply ( c ) :

b 1 ∕ n < y b − w b w + ( 1 ∕ n ) < y

for n > ( b − 1 ) ∕ ( y b − w − 1 ) . □

Proof of ( e ) . Since b w > y , b w ∕ y > 1 , and let t = b w ∕ y . Then apply ( c ) :

b w y > b 1 ∕ n b w − 1 ∕ n > y

for n > ( b − 1 ) ∕ [ ( b w ∕ y ) − 1 ] . □

Proof of ( f ) . Assume b x < y . Then, by ( b ) , there is an n > ( b − 1 ) ∕ ( y b − x − 1 ) such that b x + 1 ∕ n < y . This contradicts that x is an upper bound.

Now assume b x > y . Then, by ( c ) , there is an n > ( b − 1 ) ∕ [ ( b w ∕ y ) − 1 ] such that b x − 1 ∕ n > y . This contradicts that x is the least upper bound. Thus, b x = y . □

Proof of ( g ) . Assume there is a z ≠ x such that b x = y . If z < x , b z − x < 1 , and b z < b x . If z > x , b x − z < 1 , and b x > b z . Thus, z = x , and x is unique. □

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2023-09-01 19:10
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