Mixing
On Proving The nth odd number is 2n − 1 Through Induction, And A Few Extensions
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What I've done so far is to prove it for the n+1th number:
$(2n-1)+2=2(n+1)-1$. Because any odd number +2 is equals to the next odd number. And in the proof, it is given that 2n-1 is an odd number.
$2n+1=2n+2-1$
$2n+1=2n+1$
What I'm worried though, is if by assuming "since any odd number +2 is equals to the next odd number", am I being circular? Am I using circular reasoning?
And also, thinking about this, how would one prove that for any odd number n, the next odd number is n+2? How would they do that without being circular?
Please give the answer very simple, at the level of a pre-calculus student. I am just starting proofs for fun, and am still at a very basic level. I haven't even touched on sigma notation yet.
proof-verification induction
edited Aug 1 at 4:55
Eric Wofsey
161k12188297
asked Aug 1 at 4:16
Ethan Chan
598322
add a comment |Â
up vote
0
down vote
favorite
What I've done so far is to prove it for the n+1th number:
$(2n-1)+2=2(n+1)-1$. Because any odd number +2 is equals to the next odd number. And in the proof, it is given that 2n-1 is an odd number.
$2n+1=2n+2-1$
$2n+1=2n+1$
What I'm worried though, is if by assuming "since any odd number +2 is equals to the next odd number", am I being circular? Am I using circular reasoning?
And also, thinking about this, how would one prove that for any odd number n, the next odd number is n+2? How would they do that without being circular?
Please give the answer very simple, at the level of a pre-calculus student. I am just starting proofs for fun, and am still at a very basic level. I haven't even touched on sigma notation yet.
proof-verification induction
edited Aug 1 at 4:55
Eric Wofsey
161k12188297
asked Aug 1 at 4:16
Ethan Chan
598322
3
When you're wary about circular arguments, it's best to strip everything away and start from scratch. In particular: What is your definition of an odd number? (Also, since you're going to discuss the "$n$-th" odd number, just to be clear: Which odd number do you associate with $n=1$?)
– Blue
Aug 1 at 4:56
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
What I've done so far is to prove it for the n+1th number:
$(2n-1)+2=2(n+1)-1$. Because any odd number +2 is equals to the next odd number. And in the proof, it is given that 2n-1 is an odd number.
$2n+1=2n+2-1$
$2n+1=2n+1$
What I'm worried though, is if by assuming "since any odd number +2 is equals to the next odd number", am I being circular? Am I using circular reasoning?
And also, thinking about this, how would one prove that for any odd number n, the next odd number is n+2? How would they do that without being circular?
Please give the answer very simple, at the level of a pre-calculus student. I am just starting proofs for fun, and am still at a very basic level. I haven't even touched on sigma notation yet.
proof-verification induction
edited Aug 1 at 4:55
Eric Wofsey
161k12188297
asked Aug 1 at 4:16
Ethan Chan
598322
What I've done so far is to prove it for the n+1th number:
$(2n-1)+2=2(n+1)-1$. Because any odd number +2 is equals to the next odd number. And in the proof, it is given that 2n-1 is an odd number.
$2n+1=2n+2-1$
$2n+1=2n+1$
What I'm worried though, is if by assuming "since any odd number +2 is equals to the next odd number", am I being circular? Am I using circular reasoning?
And also, thinking about this, how would one prove that for any odd number n, the next odd number is n+2? How would they do that without being circular?
Please give the answer very simple, at the level of a pre-calculus student. I am just starting proofs for fun, and am still at a very basic level. I haven't even touched on sigma notation yet.
proof-verification induction
edited Aug 1 at 4:55
Eric Wofsey
161k12188297
asked Aug 1 at 4:16
Ethan Chan
598322
edited Aug 1 at 4:55
Eric Wofsey
161k12188297
edited Aug 1 at 4:55
Eric Wofsey
161k12188297
edited Aug 1 at 4:55
Eric Wofsey
161k12188297
161k12188297
asked Aug 1 at 4:16
Ethan Chan
598322
asked Aug 1 at 4:16
Ethan Chan
598322
asked Aug 1 at 4:16
Ethan Chan
598322
598322
3
When you're wary about circular arguments, it's best to strip everything away and start from scratch. In particular: What is your definition of an odd number? (Also, since you're going to discuss the "$n$-th" odd number, just to be clear: Which odd number do you associate with $n=1$?)
– Blue
Aug 1 at 4:56
add a comment |Â
3
When you're wary about circular arguments, it's best to strip everything away and start from scratch. In particular: What is your definition of an odd number? (Also, since you're going to discuss the "$n$-th" odd number, just to be clear: Which odd number do you associate with $n=1$?)
– Blue
Aug 1 at 4:56
3
3
When you're wary about circular arguments, it's best to strip everything away and start from scratch. In particular: What is your definition of an odd number? (Also, since you're going to discuss the "$n$-th" odd number, just to be clear: Which odd number do you associate with $n=1$?)
– Blue
Aug 1 at 4:56
When you're wary about circular arguments, it's best to strip everything away and start from scratch. In particular: What is your definition of an odd number? (Also, since you're going to discuss the "$n$-th" odd number, just to be clear: Which odd number do you associate with $n=1$?)
– Blue
Aug 1 at 4:56
add a comment |Â
2 Answers
2
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1
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accepted
$2n – 1 = 2n -2 + 1 = 2(n – 1) + 1 = 2k + 1$ where $k = n-1$ is an integer and
$ 2n -1$ is odd by definition
answered Aug 1 at 4:59
Key Flex
3,817423
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$2n - 1$ is clearly odd because $(2n - 1) textmod 2 = 1$. So $2n - 1$ is an arbitrary odd number by definition. But adding 2 results in $2n + 1$ which is odd for the same reason. Hence, adding 2 to an arbitrary odd number results in another odd number, specifically the next consecutive odd number.
answered Aug 1 at 4:51
John Reece
484
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$2n – 1 = 2n -2 + 1 = 2(n – 1) + 1 = 2k + 1$ where $k = n-1$ is an integer and
$ 2n -1$ is odd by definition
answered Aug 1 at 4:59
Key Flex
3,817423
add a comment |Â
up vote
1
down vote
accepted
$2n – 1 = 2n -2 + 1 = 2(n – 1) + 1 = 2k + 1$ where $k = n-1$ is an integer and
$ 2n -1$ is odd by definition
answered Aug 1 at 4:59
Key Flex
3,817423
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$2n – 1 = 2n -2 + 1 = 2(n – 1) + 1 = 2k + 1$ where $k = n-1$ is an integer and
$ 2n -1$ is odd by definition
answered Aug 1 at 4:59
Key Flex
3,817423
$2n – 1 = 2n -2 + 1 = 2(n – 1) + 1 = 2k + 1$ where $k = n-1$ is an integer and
$ 2n -1$ is odd by definition
answered Aug 1 at 4:59
Key Flex
3,817423
answered Aug 1 at 4:59
Key Flex
3,817423
answered Aug 1 at 4:59
Key Flex
3,817423
answered Aug 1 at 4:59
Key Flex
3,817423
3,817423
add a comment |Â
add a comment |Â
up vote
0
down vote
$2n - 1$ is clearly odd because $(2n - 1) textmod 2 = 1$. So $2n - 1$ is an arbitrary odd number by definition. But adding 2 results in $2n + 1$ which is odd for the same reason. Hence, adding 2 to an arbitrary odd number results in another odd number, specifically the next consecutive odd number.
answered Aug 1 at 4:51
John Reece
484
add a comment |Â
up vote
0
down vote
$2n - 1$ is clearly odd because $(2n - 1) textmod 2 = 1$. So $2n - 1$ is an arbitrary odd number by definition. But adding 2 results in $2n + 1$ which is odd for the same reason. Hence, adding 2 to an arbitrary odd number results in another odd number, specifically the next consecutive odd number.
answered Aug 1 at 4:51
John Reece
484
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$2n - 1$ is clearly odd because $(2n - 1) textmod 2 = 1$. So $2n - 1$ is an arbitrary odd number by definition. But adding 2 results in $2n + 1$ which is odd for the same reason. Hence, adding 2 to an arbitrary odd number results in another odd number, specifically the next consecutive odd number.
answered Aug 1 at 4:51
John Reece
484
$2n - 1$ is clearly odd because $(2n - 1) textmod 2 = 1$. So $2n - 1$ is an arbitrary odd number by definition. But adding 2 results in $2n + 1$ which is odd for the same reason. Hence, adding 2 to an arbitrary odd number results in another odd number, specifically the next consecutive odd number.
answered Aug 1 at 4:51
John Reece
484
answered Aug 1 at 4:51
John Reece
484
answered Aug 1 at 4:51
John Reece
484
answered Aug 1 at 4:51
John Reece
484
484
add a comment |Â
add a comment |Â
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3
When you're wary about circular arguments, it's best to strip everything away and start from scratch. In particular: What is your definition of an odd number? (Also, since you're going to discuss the "$n$-th" odd number, just to be clear: Which odd number do you associate with $n=1$?)
– Blue
Aug 1 at 4:56