The Real and Complex Number Systems

The Real and Complex Number Systems

Written by Men-Gen Tsai email: b89902089@ntu.edu.tw 1.

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6. Fix b > 1.

(a) If m, n, p, q are integers, n > 0, q > 0, and r = m/n = p/q, prove that

(b^m)^1/n = (b^p)^1/q. Hence it makes sense to define b^r = (b^m)^1/n. (b) Prove that b^r+s = b^rb^s if r and s are rational.

(c) If x is real, define B(x) to be the set of all numbers b^t, where t is rational and t ≤ x. Prove that

b^r = sup B(r)

where r is rational. Hence it makes sense to define b^x = sup B(x)

for every real x.

(d) Prove that b^x+y = b^xb^y for all real x and y.

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Proof: For (a): mq = np since m/n = p/q. Thus b^mq = b^np. By Theorem 1.21 we know that (b^mq)^1/(mn) = (b^np)^1/(mn), that is, (b^m)^1/n = (b^p)^1/q, that is, b^r is well-defined.

For (b): Let r = m/n and s = p/q where m, n, p, q are integers, and n > 0, q > 0. Hence (b^r+s)^nq = (b^m/n+p/q)^nq = (b(mq+np)/(nq))^nq = b^mq+np = b^mqb^np = (b^m/n)^nq(b^p/q)^nq = (b^m/nb^p/q)^nq. By Theorem 1.21 we know that ((b^r+s)^nq)^1/(nq) = ((b^m/nb^p/q)^nq)^1/(nq), that is b^r+s = b^m/nb^p/q = b^rb^s.

For (c): Note that b^r ∈ B(r). For all b^t ∈ B(r) where t is rational and t ≤ r. Hence, b^r = b^tb^r−t ≥ b^t1^r−t since b > 1 and r − t ≥ 0. Hence b^r is an upper bound of B(r). Hence b^r = sup B(r).

For (d): b^xb^y = sup B(x) sup B(y) ≥ b^txb^ty = b^tx+ty for all rational t_x ≤ x and t_y ≤ y. Note that t_x+ t_y ≤ x + y and t_x + t_y is rational.

Therefore, sup B(x) sup B(y) is a upper bound of B(x + y), that is, b^xb^y ≥ sup B(x + y) = b⁽x + y).

Conversely, we claim that b^xb^r = b^x+r if x ∈ R¹ and r ∈ Q. The following is my proof.

b^x+r = sup B(x + r) = sup{b^s : s ≤ x + r, s ∈ Q}

= sup{b^s−rb^r : s − r ≤ x, s − r ∈ Q}

= b^rsup{b^s−r : s − r ≤ x, s − r ∈ Q}

= b^rsup B(x)

= b^rb^x.

And we also claim that b^x+y ≥ b^x if y ≥ 0. The following is my proof:

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(r ∈ Q)

B(x) = {b^r : r ≤ x} ⊂ {b^r : r ≤ x + y} = B(x + y), Therefore, sup B(x + y) ≥ sup B(x), that is, b^x+y ≥ b^x. Hence,

b^x+y = sup B(x + y)

= sup{b^r : r ≤ x + y, r ∈ Q}

= sup{b^sb^r−s: r ≤ x + y, s ≤ x, r ∈ Q, s ∈ Q}

≥ sup{sup B(x)b^r−s : r ≤ x + y, s ≤ x, r ∈ Q, s ∈ Q}

= sup B(x) sup{b^r−s : r ≤ x + y, s ≤ x, r ∈ Q, s ∈ Q}

= sup B(x) sup{b^r−s : r − s ≤ x + y − s, s ≤ x, r − s ∈ Q}

= sup B(x) sup B(x + y − s)

≥ sup B(x) sup B(y)

= b^xb^y

Therefore, b^x+y = b^xb^y. 7.

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