Question

Verify the\(3x + 1\)conjecture for these integers.

a) 6

b) 7

c) 17

d) 21

Short answer

\(3x + 1\)is conjecture for 6;7;17;21 is true.

Step-by-step solution

Introduction

Let T the transformation that sends an even integer \(x\)to\(\frac{x}{2}\)and an odd integer\(x\)to\(3x + 1\).

Proving the conjecture for 6

(a)

\(\begin{aligned}{l}T(6) &= \frac{6}{2} = 3\\T(3) = 3.3 + 1 = 10\\T(10) = \frac{{10}}{2} = 5\\T(5) = 3.5 + 1 = 16\\T(16) = \frac{{16}}{2} = 8\\T(8) = \frac{8}{2} = 4\\T(4) = \frac{4}{2} = 2\\T(2) = \frac{2}{2} &= 1\end{aligned}\)

i.e.,\(6 \to 3 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1\)

Proving conjecture for 7

(b)

\(\begin{aligned}{l}T(7) &= 13.7 + 1 = 22\\T(11) = 3.11 + 1 = 34\\T(34) = \frac{{34}}{2} = 17\\T(17) = 3.17 + 1 = 52\\T(52) = \frac{{52}}{2} = 26\\T(26) = \frac{{26}}{2} = 13\\T(13) = 3.13 + 1 = 40\\T(40) = \frac{{40}}{2} = 20\\T(20) = \frac{{20}}{2} = 10\\T(10) = \frac{{10}}{2} = 5\\T(5) = 3.5 + 1 = 16\\T(16) = \frac{{16}}{2} = 8\\T(8) = \frac{8}{2} = 4\\T(4) = \frac{4}{2} = 2\\T(2) = \frac{2}{2} &= 1\end{aligned}\)

i.e.,\(7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26 \to 13 \to 40 \to 20 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1\)

Proving conjecture for 17

(c)

\(\begin{aligned}{l}T(17)& = 3.17 + 1 = 52\\T(52) = \frac{{52}}{2} = 26\\T(26) = \frac{{26}}{2} = 13\\T(13) = 3.13 + 1 = 40\\T(40) = \frac{{40}}{2} = 20\\T(20) = \frac{{20}}{2} = 10\\T(10) = \frac{{10}}{2} = 5\\T(5) = 3.5 + 1 = 16\\T(16) = \frac{{16}}{2} = 8\\T(8) = \frac{8}{2} = 4\\T(4) = \frac{4}{2} = 2\\T(2) = \frac{2}{2} &= 1\end{aligned}\)

i.e.,\(17 \to 52 \to 26 \to 13 \to 40 \to 20 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1\)

Proving conjecture for 21

(d)

\(\begin{aligned}{l}T(21) &= 3.21 + 1 = 64\\T(64) = \frac{{64}}{2} = 32\\T(32) = \frac{{32}}{2} = 16\\T(16) = \frac{{16}}{2} = 8\\T(8) = \frac{8}{2} = 4\\T(4) = \frac{4}{2} = 2\\T(2) = \frac{2}{2} &= 1\end{aligned}\)

i.e.,\(21 \to 64 \to 32 \to 16 \to 8 \to 4 \to 2 \to 1\)