Proof that an odd integer multiplied by $3$ and squared is always odd
I'm working with a proof in a discrete structures CS course, and I am a little confused by how to build up some logic for the argument. Currently we're working with symbolic logic, the problem statement is:
"For all integers n, if n is an odd integer, then $ 3n^2$ is also an odd integer."
So far I know I can build this problem by stating the two following things:
- Prove all odd numbers, multiplied by $3$ remain odd. (how can I go about proving this?)
- Prove all odd numbers square also remain odd.
What would be the best way about proving this? Would it be simpler to prove all numbers multiplied by an odd number remain odd (since an odd number squared is still being multiplied by an odd number)
I'm not really looking for a solution to be handed to me, I just feel a little stuck on where to go from here, and would like some direction on what to do next.
Thanks ahead of time guys!
elementary-number-theory proof-writing
add a comment |
I'm working with a proof in a discrete structures CS course, and I am a little confused by how to build up some logic for the argument. Currently we're working with symbolic logic, the problem statement is:
"For all integers n, if n is an odd integer, then $ 3n^2$ is also an odd integer."
So far I know I can build this problem by stating the two following things:
- Prove all odd numbers, multiplied by $3$ remain odd. (how can I go about proving this?)
- Prove all odd numbers square also remain odd.
What would be the best way about proving this? Would it be simpler to prove all numbers multiplied by an odd number remain odd (since an odd number squared is still being multiplied by an odd number)
I'm not really looking for a solution to be handed to me, I just feel a little stuck on where to go from here, and would like some direction on what to do next.
Thanks ahead of time guys!
elementary-number-theory proof-writing
add a comment |
I'm working with a proof in a discrete structures CS course, and I am a little confused by how to build up some logic for the argument. Currently we're working with symbolic logic, the problem statement is:
"For all integers n, if n is an odd integer, then $ 3n^2$ is also an odd integer."
So far I know I can build this problem by stating the two following things:
- Prove all odd numbers, multiplied by $3$ remain odd. (how can I go about proving this?)
- Prove all odd numbers square also remain odd.
What would be the best way about proving this? Would it be simpler to prove all numbers multiplied by an odd number remain odd (since an odd number squared is still being multiplied by an odd number)
I'm not really looking for a solution to be handed to me, I just feel a little stuck on where to go from here, and would like some direction on what to do next.
Thanks ahead of time guys!
elementary-number-theory proof-writing
I'm working with a proof in a discrete structures CS course, and I am a little confused by how to build up some logic for the argument. Currently we're working with symbolic logic, the problem statement is:
"For all integers n, if n is an odd integer, then $ 3n^2$ is also an odd integer."
So far I know I can build this problem by stating the two following things:
- Prove all odd numbers, multiplied by $3$ remain odd. (how can I go about proving this?)
- Prove all odd numbers square also remain odd.
What would be the best way about proving this? Would it be simpler to prove all numbers multiplied by an odd number remain odd (since an odd number squared is still being multiplied by an odd number)
I'm not really looking for a solution to be handed to me, I just feel a little stuck on where to go from here, and would like some direction on what to do next.
Thanks ahead of time guys!
elementary-number-theory proof-writing
elementary-number-theory proof-writing
edited Jan 4 at 8:49
Eevee Trainer
5,0211734
5,0211734
asked Nov 1 '14 at 20:31
dustybladedustyblade
765
765
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4 Answers
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Consider that any odd number $x$ can be written as, for some integer $n$
$$x=2n+1.$$
So, if you had, for instance two odd $x_1$ and $x_2$, you could say
$$x_1=2n_1+1$$
$$x_2=2n_2+1.$$
When you multiply these expressions, can you see how we would express the result in following form $$x_1x_2=2(text{some integer})+1$$
and hence show the product is odd?
add a comment |
The definition of "odd" is "not divisible by 2". You can see that multiplying an odd number by 3 and squaring preserves oddness by manipulating the prime factorization, which will still contain no powers of 2.
It's worth noting that this only applies to integers; by this definition, $sqrt{2}$ is odd, meaning the product of two "odd" numbers $sqrt{2}cdot sqrt{2}=2$ is even!
– Milo Brandt
Nov 1 '14 at 20:40
A positive real number has a prime factorization if and only if it is an integer (at least if we assume the exponents of the primes are nonnegative integers).
– Matt Samuel
Nov 1 '14 at 20:42
add a comment |
Any odd number $n$ can be written as $2p+1$.
$3n^2$ = $3(2p+1)^2$
=$3(4p^2+4p+1)$
The term in parenthesis is odd, since $4p^2$ and $4p$ are both even for any value of $p$.
$3$ times an odd number is odd.
add a comment |
Probably the simplest proof you can make for this:
- Let $n in mathbb{N}$ be odd.
- Thus, $2 not mid n$.
- Thus, $2 not mid n^2$.
- We also note $2 not mid 3$, and thus, $2 not mid 3n^2$.
This proof essentially makes heavy use of the fact that if $d mid ab$, then $d mid a$ or $d mid b$. The contrapositive of that claim is if $d not mid a$ and $d not mid b$, then $d not mid ab$.
(I recognize this post is old, but I felt like this proof is simpler and more intuitive, so hopefully people will appreciate this alternate proof as well.)
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Consider that any odd number $x$ can be written as, for some integer $n$
$$x=2n+1.$$
So, if you had, for instance two odd $x_1$ and $x_2$, you could say
$$x_1=2n_1+1$$
$$x_2=2n_2+1.$$
When you multiply these expressions, can you see how we would express the result in following form $$x_1x_2=2(text{some integer})+1$$
and hence show the product is odd?
add a comment |
Consider that any odd number $x$ can be written as, for some integer $n$
$$x=2n+1.$$
So, if you had, for instance two odd $x_1$ and $x_2$, you could say
$$x_1=2n_1+1$$
$$x_2=2n_2+1.$$
When you multiply these expressions, can you see how we would express the result in following form $$x_1x_2=2(text{some integer})+1$$
and hence show the product is odd?
add a comment |
Consider that any odd number $x$ can be written as, for some integer $n$
$$x=2n+1.$$
So, if you had, for instance two odd $x_1$ and $x_2$, you could say
$$x_1=2n_1+1$$
$$x_2=2n_2+1.$$
When you multiply these expressions, can you see how we would express the result in following form $$x_1x_2=2(text{some integer})+1$$
and hence show the product is odd?
Consider that any odd number $x$ can be written as, for some integer $n$
$$x=2n+1.$$
So, if you had, for instance two odd $x_1$ and $x_2$, you could say
$$x_1=2n_1+1$$
$$x_2=2n_2+1.$$
When you multiply these expressions, can you see how we would express the result in following form $$x_1x_2=2(text{some integer})+1$$
and hence show the product is odd?
answered Nov 1 '14 at 20:37
Milo BrandtMilo Brandt
39.4k475139
39.4k475139
add a comment |
add a comment |
The definition of "odd" is "not divisible by 2". You can see that multiplying an odd number by 3 and squaring preserves oddness by manipulating the prime factorization, which will still contain no powers of 2.
It's worth noting that this only applies to integers; by this definition, $sqrt{2}$ is odd, meaning the product of two "odd" numbers $sqrt{2}cdot sqrt{2}=2$ is even!
– Milo Brandt
Nov 1 '14 at 20:40
A positive real number has a prime factorization if and only if it is an integer (at least if we assume the exponents of the primes are nonnegative integers).
– Matt Samuel
Nov 1 '14 at 20:42
add a comment |
The definition of "odd" is "not divisible by 2". You can see that multiplying an odd number by 3 and squaring preserves oddness by manipulating the prime factorization, which will still contain no powers of 2.
It's worth noting that this only applies to integers; by this definition, $sqrt{2}$ is odd, meaning the product of two "odd" numbers $sqrt{2}cdot sqrt{2}=2$ is even!
– Milo Brandt
Nov 1 '14 at 20:40
A positive real number has a prime factorization if and only if it is an integer (at least if we assume the exponents of the primes are nonnegative integers).
– Matt Samuel
Nov 1 '14 at 20:42
add a comment |
The definition of "odd" is "not divisible by 2". You can see that multiplying an odd number by 3 and squaring preserves oddness by manipulating the prime factorization, which will still contain no powers of 2.
The definition of "odd" is "not divisible by 2". You can see that multiplying an odd number by 3 and squaring preserves oddness by manipulating the prime factorization, which will still contain no powers of 2.
answered Nov 1 '14 at 20:37
Matt SamuelMatt Samuel
37.4k63665
37.4k63665
It's worth noting that this only applies to integers; by this definition, $sqrt{2}$ is odd, meaning the product of two "odd" numbers $sqrt{2}cdot sqrt{2}=2$ is even!
– Milo Brandt
Nov 1 '14 at 20:40
A positive real number has a prime factorization if and only if it is an integer (at least if we assume the exponents of the primes are nonnegative integers).
– Matt Samuel
Nov 1 '14 at 20:42
add a comment |
It's worth noting that this only applies to integers; by this definition, $sqrt{2}$ is odd, meaning the product of two "odd" numbers $sqrt{2}cdot sqrt{2}=2$ is even!
– Milo Brandt
Nov 1 '14 at 20:40
A positive real number has a prime factorization if and only if it is an integer (at least if we assume the exponents of the primes are nonnegative integers).
– Matt Samuel
Nov 1 '14 at 20:42
It's worth noting that this only applies to integers; by this definition, $sqrt{2}$ is odd, meaning the product of two "odd" numbers $sqrt{2}cdot sqrt{2}=2$ is even!
– Milo Brandt
Nov 1 '14 at 20:40
It's worth noting that this only applies to integers; by this definition, $sqrt{2}$ is odd, meaning the product of two "odd" numbers $sqrt{2}cdot sqrt{2}=2$ is even!
– Milo Brandt
Nov 1 '14 at 20:40
A positive real number has a prime factorization if and only if it is an integer (at least if we assume the exponents of the primes are nonnegative integers).
– Matt Samuel
Nov 1 '14 at 20:42
A positive real number has a prime factorization if and only if it is an integer (at least if we assume the exponents of the primes are nonnegative integers).
– Matt Samuel
Nov 1 '14 at 20:42
add a comment |
Any odd number $n$ can be written as $2p+1$.
$3n^2$ = $3(2p+1)^2$
=$3(4p^2+4p+1)$
The term in parenthesis is odd, since $4p^2$ and $4p$ are both even for any value of $p$.
$3$ times an odd number is odd.
add a comment |
Any odd number $n$ can be written as $2p+1$.
$3n^2$ = $3(2p+1)^2$
=$3(4p^2+4p+1)$
The term in parenthesis is odd, since $4p^2$ and $4p$ are both even for any value of $p$.
$3$ times an odd number is odd.
add a comment |
Any odd number $n$ can be written as $2p+1$.
$3n^2$ = $3(2p+1)^2$
=$3(4p^2+4p+1)$
The term in parenthesis is odd, since $4p^2$ and $4p$ are both even for any value of $p$.
$3$ times an odd number is odd.
Any odd number $n$ can be written as $2p+1$.
$3n^2$ = $3(2p+1)^2$
=$3(4p^2+4p+1)$
The term in parenthesis is odd, since $4p^2$ and $4p$ are both even for any value of $p$.
$3$ times an odd number is odd.
answered Nov 1 '14 at 21:34
Rao A.Rao A.
1673
1673
add a comment |
add a comment |
Probably the simplest proof you can make for this:
- Let $n in mathbb{N}$ be odd.
- Thus, $2 not mid n$.
- Thus, $2 not mid n^2$.
- We also note $2 not mid 3$, and thus, $2 not mid 3n^2$.
This proof essentially makes heavy use of the fact that if $d mid ab$, then $d mid a$ or $d mid b$. The contrapositive of that claim is if $d not mid a$ and $d not mid b$, then $d not mid ab$.
(I recognize this post is old, but I felt like this proof is simpler and more intuitive, so hopefully people will appreciate this alternate proof as well.)
add a comment |
Probably the simplest proof you can make for this:
- Let $n in mathbb{N}$ be odd.
- Thus, $2 not mid n$.
- Thus, $2 not mid n^2$.
- We also note $2 not mid 3$, and thus, $2 not mid 3n^2$.
This proof essentially makes heavy use of the fact that if $d mid ab$, then $d mid a$ or $d mid b$. The contrapositive of that claim is if $d not mid a$ and $d not mid b$, then $d not mid ab$.
(I recognize this post is old, but I felt like this proof is simpler and more intuitive, so hopefully people will appreciate this alternate proof as well.)
add a comment |
Probably the simplest proof you can make for this:
- Let $n in mathbb{N}$ be odd.
- Thus, $2 not mid n$.
- Thus, $2 not mid n^2$.
- We also note $2 not mid 3$, and thus, $2 not mid 3n^2$.
This proof essentially makes heavy use of the fact that if $d mid ab$, then $d mid a$ or $d mid b$. The contrapositive of that claim is if $d not mid a$ and $d not mid b$, then $d not mid ab$.
(I recognize this post is old, but I felt like this proof is simpler and more intuitive, so hopefully people will appreciate this alternate proof as well.)
Probably the simplest proof you can make for this:
- Let $n in mathbb{N}$ be odd.
- Thus, $2 not mid n$.
- Thus, $2 not mid n^2$.
- We also note $2 not mid 3$, and thus, $2 not mid 3n^2$.
This proof essentially makes heavy use of the fact that if $d mid ab$, then $d mid a$ or $d mid b$. The contrapositive of that claim is if $d not mid a$ and $d not mid b$, then $d not mid ab$.
(I recognize this post is old, but I felt like this proof is simpler and more intuitive, so hopefully people will appreciate this alternate proof as well.)
answered Jan 4 at 8:48
Eevee TrainerEevee Trainer
5,0211734
5,0211734
add a comment |
add a comment |
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