INDUCTION why is this a valid proof? The 2019 Stack Overflow Developer Survey Results Are In Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraHow to Handle Stronger Induction Hypothesis - Strong InductionEquality such that induction step is valid but basis is not?proving $n!>2^n;;forall ;n≥4;$ by mathematical inductionMathematical Induction — $a_n=2a_n-1-1$Proof by induction that $n^3 + (n + 1)^3 + (n + 2)^3$ is a multiple of $9$. Please mark/grade.Proof by Strong Induction for $a_k = 2~a_k-1 + 3~a_k-2$Is this a valid strong induction proof? (2 base cases)Strange induction proofProof by strong induction questionIs this a valid proof that there are infinitely many natural numbers?
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INDUCTION why is this a valid proof?
The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraHow to Handle Stronger Induction Hypothesis - Strong InductionEquality such that induction step is valid but basis is not?proving $n!>2^n;;forall ;n≥4;$ by mathematical inductionMathematical Induction — $a_n=2a_n-1-1$Proof by induction that $n^3 + (n + 1)^3 + (n + 2)^3$ is a multiple of $9$. Please mark/grade.Proof by Strong Induction for $a_k = 2~a_k-1 + 3~a_k-2$Is this a valid strong induction proof? (2 base cases)Strange induction proofProof by strong induction questionIs this a valid proof that there are infinitely many natural numbers?
$begingroup$
I have a problem to understand k in the following induction proof.
Prove that $3^n -1$ is even for any natural number $n$.
Is there anybody that can show me that this is a valid proof and what is k in this case?
beginalign
tagBasis 3^0-1&=0 \
tagBasis -3^1-1&=2 \
tagInduction step textFor $ngeq2$ qquad 3^n-1&=3cdot 3^n-1-1 \
&=3cdot(3^n-1-1)+2 \
&=3cdot 2k+2 quad textwhere $k inmathbbZ$ \
&=2cdot(3k+1) \
endalign
discrete-mathematics induction proof-explanation
edited Mar 24 at 17:35
Brian
1,517416
asked Mar 24 at 16:54
user657388user657388
1
$endgroup$
|
show 7 more comments
$begingroup$
I have a problem to understand k in the following induction proof.
Prove that $3^n -1$ is even for any natural number $n$.
Is there anybody that can show me that this is a valid proof and what is k in this case?
beginalign
tagBasis 3^0-1&=0 \
tagBasis -3^1-1&=2 \
tagInduction step textFor $ngeq2$ qquad 3^n-1&=3cdot 3^n-1-1 \
&=3cdot(3^n-1-1)+2 \
&=3cdot 2k+2 quad textwhere $k inmathbbZ$ \
&=2cdot(3k+1) \
endalign
discrete-mathematics induction proof-explanation
edited Mar 24 at 17:35
Brian
1,517416
asked Mar 24 at 16:54
user657388user657388
1
$endgroup$
$begingroup$
What is it meant to be a proof by induction of?
$endgroup$
– J.G.
Mar 24 at 16:59
1
$begingroup$
Looks like you are trying to show $2|(3^n-1-1)$. If so the induction step would be assuming that there is an integer $k$ such that $2k=3^n-1-1$ with $n ge 2$
$endgroup$
– randomgirl
Mar 24 at 16:59
$begingroup$
Proof that $$3^n-1$$ is an even number for all natural numbers
$endgroup$
– user657388
Mar 24 at 17:03
$begingroup$
Proof of what? Please state what you are trying to prove.
$endgroup$
– fleablood
Mar 24 at 17:03
1
$begingroup$
But you didn't say what you are trying to prove...
$endgroup$
– CiaPan
Mar 24 at 17:07
|
show 7 more comments
$begingroup$
I have a problem to understand k in the following induction proof.
Prove that $3^n -1$ is even for any natural number $n$.
Is there anybody that can show me that this is a valid proof and what is k in this case?
beginalign
tagBasis 3^0-1&=0 \
tagBasis -3^1-1&=2 \
tagInduction step textFor $ngeq2$ qquad 3^n-1&=3cdot 3^n-1-1 \
&=3cdot(3^n-1-1)+2 \
&=3cdot 2k+2 quad textwhere $k inmathbbZ$ \
&=2cdot(3k+1) \
endalign
discrete-mathematics induction proof-explanation
edited Mar 24 at 17:35
Brian
1,517416
asked Mar 24 at 16:54
user657388user657388
1
$endgroup$
I have a problem to understand k in the following induction proof.
Prove that $3^n -1$ is even for any natural number $n$.
Is there anybody that can show me that this is a valid proof and what is k in this case?
beginalign
tagBasis 3^0-1&=0 \
tagBasis -3^1-1&=2 \
tagInduction step textFor $ngeq2$ qquad 3^n-1&=3cdot 3^n-1-1 \
&=3cdot(3^n-1-1)+2 \
&=3cdot 2k+2 quad textwhere $k inmathbbZ$ \
&=2cdot(3k+1) \
endalign
discrete-mathematics induction proof-explanation
discrete-mathematics induction proof-explanation
edited Mar 24 at 17:35
Brian
1,517416
asked Mar 24 at 16:54
user657388user657388
1
edited Mar 24 at 17:35
Brian
1,517416
asked Mar 24 at 16:54
user657388user657388
1
edited Mar 24 at 17:35
Brian
1,517416
edited Mar 24 at 17:35
Brian
1,517416
edited Mar 24 at 17:35
Brian
1,517416
1,517416
asked Mar 24 at 16:54
user657388user657388
1
asked Mar 24 at 16:54
user657388user657388
1
asked Mar 24 at 16:54
user657388user657388
1
1
$begingroup$
What is it meant to be a proof by induction of?
$endgroup$
– J.G.
Mar 24 at 16:59
1
$begingroup$
Looks like you are trying to show $2|(3^n-1-1)$. If so the induction step would be assuming that there is an integer $k$ such that $2k=3^n-1-1$ with $n ge 2$
$endgroup$
– randomgirl
Mar 24 at 16:59
$begingroup$
Proof that $$3^n-1$$ is an even number for all natural numbers
$endgroup$
– user657388
Mar 24 at 17:03
$begingroup$
Proof of what? Please state what you are trying to prove.
$endgroup$
– fleablood
Mar 24 at 17:03
1
$begingroup$
But you didn't say what you are trying to prove...
$endgroup$
– CiaPan
Mar 24 at 17:07
|
show 7 more comments
$begingroup$
What is it meant to be a proof by induction of?
$endgroup$
– J.G.
Mar 24 at 16:59
1
$begingroup$
Looks like you are trying to show $2|(3^n-1-1)$. If so the induction step would be assuming that there is an integer $k$ such that $2k=3^n-1-1$ with $n ge 2$
$endgroup$
– randomgirl
Mar 24 at 16:59
$begingroup$
Proof that $$3^n-1$$ is an even number for all natural numbers
$endgroup$
– user657388
Mar 24 at 17:03
$begingroup$
Proof of what? Please state what you are trying to prove.
$endgroup$
– fleablood
Mar 24 at 17:03
1
$begingroup$
But you didn't say what you are trying to prove...
$endgroup$
– CiaPan
Mar 24 at 17:07
$begingroup$
What is it meant to be a proof by induction of?
$endgroup$
– J.G.
Mar 24 at 16:59
$begingroup$
What is it meant to be a proof by induction of?
$endgroup$
– J.G.
Mar 24 at 16:59
1
1
$begingroup$
Looks like you are trying to show $2|(3^n-1-1)$. If so the induction step would be assuming that there is an integer $k$ such that $2k=3^n-1-1$ with $n ge 2$
$endgroup$
– randomgirl
Mar 24 at 16:59
$begingroup$
Looks like you are trying to show $2|(3^n-1-1)$. If so the induction step would be assuming that there is an integer $k$ such that $2k=3^n-1-1$ with $n ge 2$
$endgroup$
– randomgirl
Mar 24 at 16:59
$begingroup$
Proof that $$3^n-1$$ is an even number for all natural numbers
$endgroup$
– user657388
Mar 24 at 17:03
$begingroup$
Proof that $$3^n-1$$ is an even number for all natural numbers
$endgroup$
– user657388
Mar 24 at 17:03
$begingroup$
Proof of what? Please state what you are trying to prove.
$endgroup$
– fleablood
Mar 24 at 17:03
$begingroup$
Proof of what? Please state what you are trying to prove.
$endgroup$
– fleablood
Mar 24 at 17:03
1
1
$begingroup$
But you didn't say what you are trying to prove...
$endgroup$
– CiaPan
Mar 24 at 17:07
$begingroup$
But you didn't say what you are trying to prove...
$endgroup$
– CiaPan
Mar 24 at 17:07
|
show 7 more comments
2 Answers
2
active
oldest
votes
$begingroup$
I may be mistaken, but it looks like this is a proof of the fact that numbers of the form $3^n-1$ for $n in mathbbN$ are even, meaning they can be expressed as $2k$ for integral $k$. The base cases, the second of which is both superfluous and and mistakenly has a negative in front of the $3^1$, show directly that this holds for $n = 0,1$. In the inductive step, it is proven that the claim holds for $n$, assuming that it holds for $n-1$. As you can see, the quantity $3^n - 1$ is - after a string of algebraic manipulations including the substitution $3^n-1-1=2k$ are performed - shown to be equal to twice an integer. In my mind, an easier proof would be to note that $3^n$ is always odd, and thus $3^n - 1$ is always even, but maybe this is more a demonstration of the technique of induction than of its result.
edited Mar 24 at 17:08
randomgirl
3,0391915
answered Mar 24 at 17:07
Julian AsilisJulian Asilis
1
$endgroup$
add a comment |
$begingroup$
It's is a valid proof but badly explained:
Claim: $3^n -1$ is even.
Case $n = 0$. $3^0 -1 = 0$ is an even number.
Case $n = 1$ we don't actually need to do this step but $3^1 -1 = 2$ and even number. (Presumably that negative sign is a typo.)
Induction Step: $n ge 2$. AND we assume that we have already shown that $3^n-1 -1$ is an even number. (As we have already done for $n =0$ and $n=1$). [The written proof never stated that we are making that assumption. If one is familiar with induction proofs it goes without saying, but for a novice it should be pointed out.]
Now for $n$:
$3^n -1 =$
$3*3^n-1 - 1=$
$3(3^n-1 -1) + 3 -1 =$
$3(3^n-1 -1) + 2 =$
Now we assumed we have shown $3^n-1 -1$ is an even number. Let's call that numbe $2k$ and substitute it in.
$= 3(2k) + 2 = $
$6k + 2 = $
$2(3k + 1)$. ANd that is in even number.
So if $n = 2$ and we know $3^1 -1$ is even we know $3^2 - 1$ is even. ANd if $n =3$ and we know $3^2 -1$ is even we know $3^3 -1$ is even. And if $n = 4$ and we know $3^3 -1$ is even we know $3^4 -1$ is even and so on forever.
The inductive step lets us go on forever.
....
That's the basic proof by induction.
1) Base Step: You show it for the first case.
2) Induction step: You show that if it is true of one case it will be true for that next.
Implication: You realize that if one case implies the next, and you have a first case, that means you have proven the second... and therefore the third.... and therefore the fourth .... and therefore every case.
====
FWIW
I'd do it this way.
Base case: $n = 0$
$3^0 -1 = 0$ and even number.
Induction step: Assume we know that $3^n -1 = 2k$ an even number. We will prove $3^n+1 - 1$ is an even number:
$3^n+1 - 1 =$
$3*3^n - 1 = $
$3(3^n -1) + 3 - 1=$
$3*2k + 3 - 1=$
$3*2k + 2 =$
$2(3k + 1)$ is an even number.
Implication: We have shown this is true for $n=0$. We have shown if in it true for $n$ it is true for $n +1$. So it is true for $n = 0+ 1= 1$; and for $n = 1+1=2$; and for $n=2+1=3$; and so on for all non-negative integers.
answered Mar 24 at 17:23
fleabloodfleablood
1
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
I may be mistaken, but it looks like this is a proof of the fact that numbers of the form $3^n-1$ for $n in mathbbN$ are even, meaning they can be expressed as $2k$ for integral $k$. The base cases, the second of which is both superfluous and and mistakenly has a negative in front of the $3^1$, show directly that this holds for $n = 0,1$. In the inductive step, it is proven that the claim holds for $n$, assuming that it holds for $n-1$. As you can see, the quantity $3^n - 1$ is - after a string of algebraic manipulations including the substitution $3^n-1-1=2k$ are performed - shown to be equal to twice an integer. In my mind, an easier proof would be to note that $3^n$ is always odd, and thus $3^n - 1$ is always even, but maybe this is more a demonstration of the technique of induction than of its result.
edited Mar 24 at 17:08
randomgirl
3,0391915
answered Mar 24 at 17:07
Julian AsilisJulian Asilis
1
$endgroup$
add a comment |
$begingroup$
I may be mistaken, but it looks like this is a proof of the fact that numbers of the form $3^n-1$ for $n in mathbbN$ are even, meaning they can be expressed as $2k$ for integral $k$. The base cases, the second of which is both superfluous and and mistakenly has a negative in front of the $3^1$, show directly that this holds for $n = 0,1$. In the inductive step, it is proven that the claim holds for $n$, assuming that it holds for $n-1$. As you can see, the quantity $3^n - 1$ is - after a string of algebraic manipulations including the substitution $3^n-1-1=2k$ are performed - shown to be equal to twice an integer. In my mind, an easier proof would be to note that $3^n$ is always odd, and thus $3^n - 1$ is always even, but maybe this is more a demonstration of the technique of induction than of its result.
edited Mar 24 at 17:08
randomgirl
3,0391915
answered Mar 24 at 17:07
Julian AsilisJulian Asilis
1
$endgroup$
add a comment |
$begingroup$
I may be mistaken, but it looks like this is a proof of the fact that numbers of the form $3^n-1$ for $n in mathbbN$ are even, meaning they can be expressed as $2k$ for integral $k$. The base cases, the second of which is both superfluous and and mistakenly has a negative in front of the $3^1$, show directly that this holds for $n = 0,1$. In the inductive step, it is proven that the claim holds for $n$, assuming that it holds for $n-1$. As you can see, the quantity $3^n - 1$ is - after a string of algebraic manipulations including the substitution $3^n-1-1=2k$ are performed - shown to be equal to twice an integer. In my mind, an easier proof would be to note that $3^n$ is always odd, and thus $3^n - 1$ is always even, but maybe this is more a demonstration of the technique of induction than of its result.
edited Mar 24 at 17:08
randomgirl
3,0391915
answered Mar 24 at 17:07
Julian AsilisJulian Asilis
1
$endgroup$
I may be mistaken, but it looks like this is a proof of the fact that numbers of the form $3^n-1$ for $n in mathbbN$ are even, meaning they can be expressed as $2k$ for integral $k$. The base cases, the second of which is both superfluous and and mistakenly has a negative in front of the $3^1$, show directly that this holds for $n = 0,1$. In the inductive step, it is proven that the claim holds for $n$, assuming that it holds for $n-1$. As you can see, the quantity $3^n - 1$ is - after a string of algebraic manipulations including the substitution $3^n-1-1=2k$ are performed - shown to be equal to twice an integer. In my mind, an easier proof would be to note that $3^n$ is always odd, and thus $3^n - 1$ is always even, but maybe this is more a demonstration of the technique of induction than of its result.
edited Mar 24 at 17:08
randomgirl
3,0391915
answered Mar 24 at 17:07
Julian AsilisJulian Asilis
1
edited Mar 24 at 17:08
randomgirl
3,0391915
edited Mar 24 at 17:08
randomgirl
3,0391915
edited Mar 24 at 17:08
randomgirl
3,0391915
3,0391915
answered Mar 24 at 17:07
Julian AsilisJulian Asilis
1
answered Mar 24 at 17:07
Julian AsilisJulian Asilis
1
answered Mar 24 at 17:07
Julian AsilisJulian Asilis
1
1
add a comment |
add a comment |
$begingroup$
It's is a valid proof but badly explained:
Claim: $3^n -1$ is even.
Case $n = 0$. $3^0 -1 = 0$ is an even number.
Case $n = 1$ we don't actually need to do this step but $3^1 -1 = 2$ and even number. (Presumably that negative sign is a typo.)
Induction Step: $n ge 2$. AND we assume that we have already shown that $3^n-1 -1$ is an even number. (As we have already done for $n =0$ and $n=1$). [The written proof never stated that we are making that assumption. If one is familiar with induction proofs it goes without saying, but for a novice it should be pointed out.]
Now for $n$:
$3^n -1 =$
$3*3^n-1 - 1=$
$3(3^n-1 -1) + 3 -1 =$
$3(3^n-1 -1) + 2 =$
Now we assumed we have shown $3^n-1 -1$ is an even number. Let's call that numbe $2k$ and substitute it in.
$= 3(2k) + 2 = $
$6k + 2 = $
$2(3k + 1)$. ANd that is in even number.
So if $n = 2$ and we know $3^1 -1$ is even we know $3^2 - 1$ is even. ANd if $n =3$ and we know $3^2 -1$ is even we know $3^3 -1$ is even. And if $n = 4$ and we know $3^3 -1$ is even we know $3^4 -1$ is even and so on forever.
The inductive step lets us go on forever.
....
That's the basic proof by induction.
1) Base Step: You show it for the first case.
2) Induction step: You show that if it is true of one case it will be true for that next.
Implication: You realize that if one case implies the next, and you have a first case, that means you have proven the second... and therefore the third.... and therefore the fourth .... and therefore every case.
====
FWIW
I'd do it this way.
Base case: $n = 0$
$3^0 -1 = 0$ and even number.
Induction step: Assume we know that $3^n -1 = 2k$ an even number. We will prove $3^n+1 - 1$ is an even number:
$3^n+1 - 1 =$
$3*3^n - 1 = $
$3(3^n -1) + 3 - 1=$
$3*2k + 3 - 1=$
$3*2k + 2 =$
$2(3k + 1)$ is an even number.
Implication: We have shown this is true for $n=0$. We have shown if in it true for $n$ it is true for $n +1$. So it is true for $n = 0+ 1= 1$; and for $n = 1+1=2$; and for $n=2+1=3$; and so on for all non-negative integers.
answered Mar 24 at 17:23
fleabloodfleablood
1
$endgroup$
add a comment |
$begingroup$
It's is a valid proof but badly explained:
Claim: $3^n -1$ is even.
Case $n = 0$. $3^0 -1 = 0$ is an even number.
Case $n = 1$ we don't actually need to do this step but $3^1 -1 = 2$ and even number. (Presumably that negative sign is a typo.)
Induction Step: $n ge 2$. AND we assume that we have already shown that $3^n-1 -1$ is an even number. (As we have already done for $n =0$ and $n=1$). [The written proof never stated that we are making that assumption. If one is familiar with induction proofs it goes without saying, but for a novice it should be pointed out.]
Now for $n$:
$3^n -1 =$
$3*3^n-1 - 1=$
$3(3^n-1 -1) + 3 -1 =$
$3(3^n-1 -1) + 2 =$
Now we assumed we have shown $3^n-1 -1$ is an even number. Let's call that numbe $2k$ and substitute it in.
$= 3(2k) + 2 = $
$6k + 2 = $
$2(3k + 1)$. ANd that is in even number.
So if $n = 2$ and we know $3^1 -1$ is even we know $3^2 - 1$ is even. ANd if $n =3$ and we know $3^2 -1$ is even we know $3^3 -1$ is even. And if $n = 4$ and we know $3^3 -1$ is even we know $3^4 -1$ is even and so on forever.
The inductive step lets us go on forever.
....
That's the basic proof by induction.
1) Base Step: You show it for the first case.
2) Induction step: You show that if it is true of one case it will be true for that next.
Implication: You realize that if one case implies the next, and you have a first case, that means you have proven the second... and therefore the third.... and therefore the fourth .... and therefore every case.
====
FWIW
I'd do it this way.
Base case: $n = 0$
$3^0 -1 = 0$ and even number.
Induction step: Assume we know that $3^n -1 = 2k$ an even number. We will prove $3^n+1 - 1$ is an even number:
$3^n+1 - 1 =$
$3*3^n - 1 = $
$3(3^n -1) + 3 - 1=$
$3*2k + 3 - 1=$
$3*2k + 2 =$
$2(3k + 1)$ is an even number.
Implication: We have shown this is true for $n=0$. We have shown if in it true for $n$ it is true for $n +1$. So it is true for $n = 0+ 1= 1$; and for $n = 1+1=2$; and for $n=2+1=3$; and so on for all non-negative integers.
answered Mar 24 at 17:23
fleabloodfleablood
1
$endgroup$
add a comment |
$begingroup$
It's is a valid proof but badly explained:
Claim: $3^n -1$ is even.
Case $n = 0$. $3^0 -1 = 0$ is an even number.
Case $n = 1$ we don't actually need to do this step but $3^1 -1 = 2$ and even number. (Presumably that negative sign is a typo.)
Induction Step: $n ge 2$. AND we assume that we have already shown that $3^n-1 -1$ is an even number. (As we have already done for $n =0$ and $n=1$). [The written proof never stated that we are making that assumption. If one is familiar with induction proofs it goes without saying, but for a novice it should be pointed out.]
Now for $n$:
$3^n -1 =$
$3*3^n-1 - 1=$
$3(3^n-1 -1) + 3 -1 =$
$3(3^n-1 -1) + 2 =$
Now we assumed we have shown $3^n-1 -1$ is an even number. Let's call that numbe $2k$ and substitute it in.
$= 3(2k) + 2 = $
$6k + 2 = $
$2(3k + 1)$. ANd that is in even number.
So if $n = 2$ and we know $3^1 -1$ is even we know $3^2 - 1$ is even. ANd if $n =3$ and we know $3^2 -1$ is even we know $3^3 -1$ is even. And if $n = 4$ and we know $3^3 -1$ is even we know $3^4 -1$ is even and so on forever.
The inductive step lets us go on forever.
....
That's the basic proof by induction.
1) Base Step: You show it for the first case.
2) Induction step: You show that if it is true of one case it will be true for that next.
Implication: You realize that if one case implies the next, and you have a first case, that means you have proven the second... and therefore the third.... and therefore the fourth .... and therefore every case.
====
FWIW
I'd do it this way.
Base case: $n = 0$
$3^0 -1 = 0$ and even number.
Induction step: Assume we know that $3^n -1 = 2k$ an even number. We will prove $3^n+1 - 1$ is an even number:
$3^n+1 - 1 =$
$3*3^n - 1 = $
$3(3^n -1) + 3 - 1=$
$3*2k + 3 - 1=$
$3*2k + 2 =$
$2(3k + 1)$ is an even number.
Implication: We have shown this is true for $n=0$. We have shown if in it true for $n$ it is true for $n +1$. So it is true for $n = 0+ 1= 1$; and for $n = 1+1=2$; and for $n=2+1=3$; and so on for all non-negative integers.
answered Mar 24 at 17:23
fleabloodfleablood
1
$endgroup$
It's is a valid proof but badly explained:
Claim: $3^n -1$ is even.
Case $n = 0$. $3^0 -1 = 0$ is an even number.
Case $n = 1$ we don't actually need to do this step but $3^1 -1 = 2$ and even number. (Presumably that negative sign is a typo.)
Induction Step: $n ge 2$. AND we assume that we have already shown that $3^n-1 -1$ is an even number. (As we have already done for $n =0$ and $n=1$). [The written proof never stated that we are making that assumption. If one is familiar with induction proofs it goes without saying, but for a novice it should be pointed out.]
Now for $n$:
$3^n -1 =$
$3*3^n-1 - 1=$
$3(3^n-1 -1) + 3 -1 =$
$3(3^n-1 -1) + 2 =$
Now we assumed we have shown $3^n-1 -1$ is an even number. Let's call that numbe $2k$ and substitute it in.
$= 3(2k) + 2 = $
$6k + 2 = $
$2(3k + 1)$. ANd that is in even number.
So if $n = 2$ and we know $3^1 -1$ is even we know $3^2 - 1$ is even. ANd if $n =3$ and we know $3^2 -1$ is even we know $3^3 -1$ is even. And if $n = 4$ and we know $3^3 -1$ is even we know $3^4 -1$ is even and so on forever.
The inductive step lets us go on forever.
....
That's the basic proof by induction.
1) Base Step: You show it for the first case.
2) Induction step: You show that if it is true of one case it will be true for that next.
Implication: You realize that if one case implies the next, and you have a first case, that means you have proven the second... and therefore the third.... and therefore the fourth .... and therefore every case.
====
FWIW
I'd do it this way.
Base case: $n = 0$
$3^0 -1 = 0$ and even number.
Induction step: Assume we know that $3^n -1 = 2k$ an even number. We will prove $3^n+1 - 1$ is an even number:
$3^n+1 - 1 =$
$3*3^n - 1 = $
$3(3^n -1) + 3 - 1=$
$3*2k + 3 - 1=$
$3*2k + 2 =$
$2(3k + 1)$ is an even number.
Implication: We have shown this is true for $n=0$. We have shown if in it true for $n$ it is true for $n +1$. So it is true for $n = 0+ 1= 1$; and for $n = 1+1=2$; and for $n=2+1=3$; and so on for all non-negative integers.
answered Mar 24 at 17:23
fleabloodfleablood
1
edited Mar 24 at 17:29
answered Mar 24 at 17:23
fleabloodfleablood
1
answered Mar 24 at 17:23
fleabloodfleablood
1
answered Mar 24 at 17:23
fleabloodfleablood
1
1
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$begingroup$
What is it meant to be a proof by induction of?
$endgroup$
– J.G.
Mar 24 at 16:59
1
$begingroup$
Looks like you are trying to show $2|(3^n-1-1)$. If so the induction step would be assuming that there is an integer $k$ such that $2k=3^n-1-1$ with $n ge 2$
$endgroup$
– randomgirl
Mar 24 at 16:59
$begingroup$
Proof that $$3^n-1$$ is an even number for all natural numbers
$endgroup$
– user657388
Mar 24 at 17:03
$begingroup$
Proof of what? Please state what you are trying to prove.
$endgroup$
– fleablood
Mar 24 at 17:03
1
$begingroup$
But you didn't say what you are trying to prove...
$endgroup$
– CiaPan
Mar 24 at 17:07