Proof by induction

Randyyy

New member
Joined
May 14, 2020
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Hey, Mathhelp! Is my proof valid? I am new to induction so I am not too confident what I have done is correct.


Prove that 3n-1 is divisible by 2 \(\displaystyle \forall n \in\mathbb{Z}^+ ,n\geq 1\)
S(n)=3n-1=2k, \(\displaystyle k\in\mathbb{Z}\)
Base-case n=1: 31-1=2*1, true for n=1
S(p)=3p-1=2k
Assuming it is true for n=p, then it is true for n=p+1.
S(p+1)= 3p+1-1=2r, \(\displaystyle r\in\mathbb{Z}\)
S(p)=3(3p-1)=3*2k
S(p)=3p+1-3=6k
S(p)=3p+1-1=6k+2
S(p)=3p+1-1=2(3k+1)
(3k+1)=r and hence:
S(p)=3p+1-1=2r
 

pka

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Jan 29, 2005
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10,453
Prove that 3n-1 is divisible by 2 \(\displaystyle \forall n \in\mathbb{Z}^+ ,n\geq 1\)
Lets's suppose that \(3^p-1=2k\) where \(\{p,k\}\subset\mathbb{Z}^+\)
Look at
\( \begin{align*}3^{p+1}-1&=3\cdot 3^p-1 \\&=3\cdot 3^p-3+2\\&=3\cdot (3^p-1)+2\\&=3\cdot (2k)+2\\&=2(3k+1)\text{ which is even} \end{align*}\)
 
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Randyyy

New member
Joined
May 14, 2020
Messages
43
Lets's suppose that \(3^p-1=2k\) where \(\{p,k\}\subset\mathbb{Z}^+\)
Look at
\( \begin{align*}3^{p+1}-1&=3\cdot 3^p-1 \\&=3\cdot 3^p-3+2\\&=3\cdot (3^p-1)+2\\&=3\cdot (2k)+2\\&=2(3k+1)\text{ which is even} \end{align*}\)
Thank you
 
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