Question

Define the exponentiation operator on naturals recursively so that x^0 = 1 and x^(S(y)) = x^y * x. Prove by induction, using this definition, that for any naturals x, y, and z, x^(y+z) = x^y * x^z and x^(y*z) = (x^y)^z.

Name: define the exponentiation operator on naturals recursively so that x0 1 and xsy xy x prove by induction using this definition that for any naturals x y and z xyz xy xz and xyz xy 85482
Uploaded: 2024-03-30T21:55:44-08:00
Duration: 2 min 17 s
Channel: Hoan Nguyen
Description: define the exponentiation operator on naturals recursively so that x0 1 and xsy xy x prove by induction using this definition that for any naturals x y and z xyz xy xz and xyz xy 85482

          Define the exponentiation operator on naturals recursively so that x^0 = 1 and x^(S(y)) = x^y * x. Prove by induction, using this definition, that for any naturals x, y, and z, x^(y+z) = x^y * x^z and x^(y*z) = (x^y)^z.

Added by Judy W.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Hoan Nguyen

03/30/2024

Step 1

- Base case: \(x^0 = 1\) - Recursive step: \(x^{S(y)} = x^y * x\), where \(S(y)\) denotes the successor of \(y\), i.e., \(y + 1\). Show more…

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Define the exponentiation operator on naturals recursively so that x^0 = 1 and x^(S(y)) = x^y * x. Prove by induction, using this definition, that for any naturals x, y, and z, x^(y+z) = x^y * x^z and x^(y*z) = (x^y)^z.