2. Prove or disprove: for any real number x there is a unique map $a: \mathbb{Z} \to \{-1, 0, 1\}$ such that $x = \sum_{n \in \mathbb{Z}} a_n 3^n$, for any $m \in \mathbb{Z}$ there is $n \le m$ such that $a_n \ne -1$, and there is $m \in \mathbb{Z}$ such that for any $n \ge m$ we have $a_n = 0$.

          2. Prove or disprove: for any real number x there is a unique map $a: \mathbb{Z} \to \{-1, 0, 1\}$ such that
$x = \sum_{n \in \mathbb{Z}} a_n 3^n$,
for any $m \in \mathbb{Z}$ there is $n \le m$ such that $a_n \ne -1$, and there is $m \in \mathbb{Z}$ such that for any $n \ge m$ we have
$a_n = 0$.

$2. Prove or disprove: for any real number x there is a unique map a: ℤ→{-1, 0, 1} such that x = ∑n ∈ℤ an 3^n, for any m ∈ℤ there is n ≤ m such that an -1, and there is m ∈ℤ such that for any n ≥ m we have an = 0.$

Added by Claudia E.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert James Chok

11/26/2019

Step 1

Step 1: Consider the first statement: for any real number x there is a unique map a: Z -> {-1,0,1} such that x=Σ_(n∈Z) a_n3^n. Show more…

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Please help, I can't seem to get this one. Prove or disprove: for any real number x there is a unique map a: Z -> {-1,0,1} such that x=Σ_(n∈Z) a_n3^n for any m∈Z there is n≤m such that a_n≠-1, and there is m∈Z such that for any n≥m we have a_n=0. 2. Prove or disprove: for any real number x there is a unique map a: Z -> {-1,0,1} such that x=Σ_(n∈Z) a_n3^n for any m∈Z there is n<m such that a_n≠-1, and there is m∈Z such that for any n≥m we have a_n=0.