Question

The function $h: \mathbb{Z}_{12} \rightarrow \mathbb{Z}_{18}$ is defined by the formula $h([a]) = [3a - 5]$. (a) Verify that the function $h$ is well-defined, that is, show that if $a, b \in \mathbb{Z}$ are such that $[a] = [b]$ in $\mathbb{Z}_{12}$ then $[3a - 5] = [3b - 5]$ in $\mathbb{Z}_{18}$.

          The function $h: \mathbb{Z}_{12} \rightarrow \mathbb{Z}_{18}$ is defined by the formula $h([a]) = [3a - 5]$.
(a) Verify that the function $h$ is well-defined, that is, show that if $a, b \in \mathbb{Z}$ are such that $[a] = [b]$ in $\mathbb{Z}_{12}$ then $[3a - 5] = [3b - 5]$ in $\mathbb{Z}_{18}$.

The function h: ℤ12→ℤ18 is defined by the formula h([a]) = [3a - 5].
(a) Verify that the function h is well-defined, that is, show that if a, b ∈ℤ are such that [a] = [b] in ℤ12 then [3a - 5] = [3b - 5] in ℤ18.

Added by Christina C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert James Kiss

07/20/2020

Step 1

Step 1: If [a] = [b] in Z12, then a ≡ b (mod 12). Show more…

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The function h:Z_(12)->Z_(18) is defined by the formula h([a])=[3a-5]. (a) Verify that the function h is well-defined, that is, show that if a,binZ are such that [a]=[b] in Z_(12) then [3a-5]=[3b-5] in Z_(18). Proof please The function h : Z12 -> Z18 is defined by the formula h([a]) = [3a - 5]. (a) Verify that the function h is well-defined, that is, show that if a, b E Z are such that [a] = [b in Z12 then [3a- 5]= [3b- 5] in Z18