Mixing
Simple practice proof
Clash Royale CLAN TAG#URR8PPP
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I am still trying to master the notion of proofs. It is quite a difficult transition, and I am not the best at it. Any suggestions or corrections on my logic would be extremely helpful. Thanks in advance.
Claim: If $n$ is an integer, then $n^2 geq n$.
I rewrote this as:
For all integers $n$, $n^2 geq n$.
Proof: Suppose $n^2$ < $n$, for all integers $n$. Then, $n cdot n < n implies n < 1$. This is not true since $n in mathbbZ$ and $n$ can also be greater than or equal to $1$. Thus, contradicting our assumption.
I feel like I completely butchered this proof, but I am just starting to practice it more. Any help would be greatly appreciated.
number-theory
edited Jul 27 at 2:44
Math Lover
12.3k21232
asked Jul 27 at 2:40
Ryan
865
add a comment |Â
up vote
1
down vote
favorite
I am still trying to master the notion of proofs. It is quite a difficult transition, and I am not the best at it. Any suggestions or corrections on my logic would be extremely helpful. Thanks in advance.
Claim: If $n$ is an integer, then $n^2 geq n$.
I rewrote this as:
For all integers $n$, $n^2 geq n$.
Proof: Suppose $n^2$ < $n$, for all integers $n$. Then, $n cdot n < n implies n < 1$. This is not true since $n in mathbbZ$ and $n$ can also be greater than or equal to $1$. Thus, contradicting our assumption.
I feel like I completely butchered this proof, but I am just starting to practice it more. Any help would be greatly appreciated.
number-theory
edited Jul 27 at 2:44
Math Lover
12.3k21232
asked Jul 27 at 2:40
Ryan
865
I don't see the need for proof by contradiction.
– Shrey Joshi
Jul 27 at 2:47
If you do a proof by contradiction (not recommended, but that seems to be how you are starting), the negation of: 'For all integers $n$, $n^2>n$' is not 'For all integers $n$, $n^2<n$' ; but rather: 'There exists an integer $n$ such that $n^2<n$. That is, the negation of a universal statement (for all), is an existential statement (there exists)..
– paw88789
Jul 27 at 2:53
Thanks. Would a direct proof be adequate? If so, how do I go about it? Intuitively, I can look at the claim and know it is true, but writing it in proof format is the hard part.
– Ryan
Jul 27 at 2:56
1
Both direct and indirect have the same principal. If $n ge 1$ then $n*n ge n*1$ and that works whether you divide to go from $n^2 < n implies n < 1$ which is a contradiction or if you go from $n ge 1 implies n^2 ge n$. But what would you do if some pick pendant (like me) comes along and asks "Why do you say $ncdot n < n$ would imply $n < 1$? Why do you claim that? How do you know it is true?" Would you have an answer?
– fleablood
Jul 27 at 3:12
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am still trying to master the notion of proofs. It is quite a difficult transition, and I am not the best at it. Any suggestions or corrections on my logic would be extremely helpful. Thanks in advance.
Claim: If $n$ is an integer, then $n^2 geq n$.
I rewrote this as:
For all integers $n$, $n^2 geq n$.
Proof: Suppose $n^2$ < $n$, for all integers $n$. Then, $n cdot n < n implies n < 1$. This is not true since $n in mathbbZ$ and $n$ can also be greater than or equal to $1$. Thus, contradicting our assumption.
I feel like I completely butchered this proof, but I am just starting to practice it more. Any help would be greatly appreciated.
number-theory
edited Jul 27 at 2:44
Math Lover
12.3k21232
asked Jul 27 at 2:40
Ryan
865
I am still trying to master the notion of proofs. It is quite a difficult transition, and I am not the best at it. Any suggestions or corrections on my logic would be extremely helpful. Thanks in advance.
Claim: If $n$ is an integer, then $n^2 geq n$.
I rewrote this as:
For all integers $n$, $n^2 geq n$.
Proof: Suppose $n^2$ < $n$, for all integers $n$. Then, $n cdot n < n implies n < 1$. This is not true since $n in mathbbZ$ and $n$ can also be greater than or equal to $1$. Thus, contradicting our assumption.
I feel like I completely butchered this proof, but I am just starting to practice it more. Any help would be greatly appreciated.
number-theory
edited Jul 27 at 2:44
Math Lover
12.3k21232
asked Jul 27 at 2:40
Ryan
865
edited Jul 27 at 2:44
Math Lover
12.3k21232
edited Jul 27 at 2:44
Math Lover
12.3k21232
edited Jul 27 at 2:44
Math Lover
12.3k21232
12.3k21232
asked Jul 27 at 2:40
Ryan
865
asked Jul 27 at 2:40
Ryan
865
asked Jul 27 at 2:40
Ryan
865
865
I don't see the need for proof by contradiction.
– Shrey Joshi
Jul 27 at 2:47
If you do a proof by contradiction (not recommended, but that seems to be how you are starting), the negation of: 'For all integers $n$, $n^2>n$' is not 'For all integers $n$, $n^2<n$' ; but rather: 'There exists an integer $n$ such that $n^2<n$. That is, the negation of a universal statement (for all), is an existential statement (there exists)..
– paw88789
Jul 27 at 2:53
Thanks. Would a direct proof be adequate? If so, how do I go about it? Intuitively, I can look at the claim and know it is true, but writing it in proof format is the hard part.
– Ryan
Jul 27 at 2:56
1
Both direct and indirect have the same principal. If $n ge 1$ then $n*n ge n*1$ and that works whether you divide to go from $n^2 < n implies n < 1$ which is a contradiction or if you go from $n ge 1 implies n^2 ge n$. But what would you do if some pick pendant (like me) comes along and asks "Why do you say $ncdot n < n$ would imply $n < 1$? Why do you claim that? How do you know it is true?" Would you have an answer?
– fleablood
Jul 27 at 3:12
add a comment |Â
I don't see the need for proof by contradiction.
– Shrey Joshi
Jul 27 at 2:47
If you do a proof by contradiction (not recommended, but that seems to be how you are starting), the negation of: 'For all integers $n$, $n^2>n$' is not 'For all integers $n$, $n^2<n$' ; but rather: 'There exists an integer $n$ such that $n^2<n$. That is, the negation of a universal statement (for all), is an existential statement (there exists)..
– paw88789
Jul 27 at 2:53
Thanks. Would a direct proof be adequate? If so, how do I go about it? Intuitively, I can look at the claim and know it is true, but writing it in proof format is the hard part.
– Ryan
Jul 27 at 2:56
1
Both direct and indirect have the same principal. If $n ge 1$ then $n*n ge n*1$ and that works whether you divide to go from $n^2 < n implies n < 1$ which is a contradiction or if you go from $n ge 1 implies n^2 ge n$. But what would you do if some pick pendant (like me) comes along and asks "Why do you say $ncdot n < n$ would imply $n < 1$? Why do you claim that? How do you know it is true?" Would you have an answer?
– fleablood
Jul 27 at 3:12
I don't see the need for proof by contradiction.
– Shrey Joshi
Jul 27 at 2:47
I don't see the need for proof by contradiction.
– Shrey Joshi
Jul 27 at 2:47
If you do a proof by contradiction (not recommended, but that seems to be how you are starting), the negation of: 'For all integers $n$, $n^2>n$' is not 'For all integers $n$, $n^2<n$' ; but rather: 'There exists an integer $n$ such that $n^2<n$. That is, the negation of a universal statement (for all), is an existential statement (there exists)..
– paw88789
Jul 27 at 2:53
If you do a proof by contradiction (not recommended, but that seems to be how you are starting), the negation of: 'For all integers $n$, $n^2>n$' is not 'For all integers $n$, $n^2<n$' ; but rather: 'There exists an integer $n$ such that $n^2<n$. That is, the negation of a universal statement (for all), is an existential statement (there exists)..
– paw88789
Jul 27 at 2:53
Thanks. Would a direct proof be adequate? If so, how do I go about it? Intuitively, I can look at the claim and know it is true, but writing it in proof format is the hard part.
– Ryan
Jul 27 at 2:56
Thanks. Would a direct proof be adequate? If so, how do I go about it? Intuitively, I can look at the claim and know it is true, but writing it in proof format is the hard part.
– Ryan
Jul 27 at 2:56
1
1
Both direct and indirect have the same principal. If $n ge 1$ then $n*n ge n*1$ and that works whether you divide to go from $n^2 < n implies n < 1$ which is a contradiction or if you go from $n ge 1 implies n^2 ge n$. But what would you do if some pick pendant (like me) comes along and asks "Why do you say $ncdot n < n$ would imply $n < 1$? Why do you claim that? How do you know it is true?" Would you have an answer?
– fleablood
Jul 27 at 3:12
Both direct and indirect have the same principal. If $n ge 1$ then $n*n ge n*1$ and that works whether you divide to go from $n^2 < n implies n < 1$ which is a contradiction or if you go from $n ge 1 implies n^2 ge n$. But what would you do if some pick pendant (like me) comes along and asks "Why do you say $ncdot n < n$ would imply $n < 1$? Why do you claim that? How do you know it is true?" Would you have an answer?
– fleablood
Jul 27 at 3:12
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
0
down vote
Your proof is fine, after the modification in the comments. However, you can do a direct proof also:
Let $n$ be any integer. Thus $n ge 1$. Multiplying this inequality* with $n ge n$ we get $n^2 ge n$.
*In the definition of $ge$ it is said that if $a ge b$ and $c ge d$, and all quantities are positive, then $ac ge bd$.
answered Jul 27 at 2:57
Ovi
11.3k935105
Thank you for your feedback!
– Ryan
Jul 27 at 3:02
1
@Ryan You're welcome!
– Ovi
Jul 27 at 3:03
add a comment |Â
up vote
0
down vote
Claim: $n^2geq n$ for all $ninmathbbZ$.
If there is even a single counterexample, then we know that the claim is false. So in fact, the true negation of the claim is:
Suppose $n^2<n$ for some $ninmathbbZ$.
The rest of the proof follows like this:
If $n^2<n$ for some $ninmathbbZ$, then $n^2-n<0implies n(n-1)<0$.
Consider the following three cases:
$1)$ If $n=0$, then $n(n-1)=0nless0$ so we arrive at a contradiction.
$2)$ If $n>0$, then $n-1geq 0$ so $n(n-1)nless 0$ and we arrive at a contradiction.
$3)$ If $n<0$, then $n-1<0$ so $n(n-1)>0$ and we arrive at a contradiction.
QED
answered Jul 27 at 2:58
高田航
1,116318
1
Thank you for breaking it down so easily for me to understand.
– Ryan
Jul 27 at 3:02
add a comment |Â
up vote
0
down vote
The negation of "for all" is "there exists". If you want to proceed solving this question with a proof by contradiction, your starting assumption must be "Suppose $n^2<n$ for some integer $nin mathbbZ$.
Then, your need to be a little more careful in the writing of the proof. Indeed, you are trying to divide your inequality by $n$ to deduce a contradiction, but there are two things you need to be attentive to. First, $n$ must not be zero. The case $n=0$, which is trivial, must be mentioned separatedly. Second, if $n$ is negative, then the order of the inequality changes. So you need to treat the cases $n$ positive and $n$ negative separatedly (and conclude as you did).
Now, because of these cases that need to be distinguished, this kind of proof is not recommended.
You may show the result directly, by noticing that $n^2 geq n$ if and only if $n(n-1)geq 0$. A product of two quantities is non-negative if and only if both have the same sign. Thus, you are reduced to proving that for every $n in mathbbZ$, $n$ and $n-1$ have the same sign (with the meaning "both are non-negative or both are non-positive"). This is now rather straightforward.
answered Jul 27 at 2:56
Suzet
2,203427
Gotcha! Thank you so much for your help.
– Ryan
Jul 27 at 2:57
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Your proof is fine, after the modification in the comments. However, you can do a direct proof also:
Let $n$ be any integer. Thus $n ge 1$. Multiplying this inequality* with $n ge n$ we get $n^2 ge n$.
*In the definition of $ge$ it is said that if $a ge b$ and $c ge d$, and all quantities are positive, then $ac ge bd$.
answered Jul 27 at 2:57
Ovi
11.3k935105
Thank you for your feedback!
– Ryan
Jul 27 at 3:02
1
@Ryan You're welcome!
– Ovi
Jul 27 at 3:03
add a comment |Â
up vote
0
down vote
Your proof is fine, after the modification in the comments. However, you can do a direct proof also:
Let $n$ be any integer. Thus $n ge 1$. Multiplying this inequality* with $n ge n$ we get $n^2 ge n$.
*In the definition of $ge$ it is said that if $a ge b$ and $c ge d$, and all quantities are positive, then $ac ge bd$.
answered Jul 27 at 2:57
Ovi
11.3k935105
Thank you for your feedback!
– Ryan
Jul 27 at 3:02
1
@Ryan You're welcome!
– Ovi
Jul 27 at 3:03
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your proof is fine, after the modification in the comments. However, you can do a direct proof also:
Let $n$ be any integer. Thus $n ge 1$. Multiplying this inequality* with $n ge n$ we get $n^2 ge n$.
*In the definition of $ge$ it is said that if $a ge b$ and $c ge d$, and all quantities are positive, then $ac ge bd$.
answered Jul 27 at 2:57
Ovi
11.3k935105
Your proof is fine, after the modification in the comments. However, you can do a direct proof also:
Let $n$ be any integer. Thus $n ge 1$. Multiplying this inequality* with $n ge n$ we get $n^2 ge n$.
*In the definition of $ge$ it is said that if $a ge b$ and $c ge d$, and all quantities are positive, then $ac ge bd$.
answered Jul 27 at 2:57
Ovi
11.3k935105
answered Jul 27 at 2:57
Ovi
11.3k935105
answered Jul 27 at 2:57
Ovi
11.3k935105
answered Jul 27 at 2:57
Ovi
11.3k935105
11.3k935105
Thank you for your feedback!
– Ryan
Jul 27 at 3:02
1
@Ryan You're welcome!
– Ovi
Jul 27 at 3:03
add a comment |Â
Thank you for your feedback!
– Ryan
Jul 27 at 3:02
1
@Ryan You're welcome!
– Ovi
Jul 27 at 3:03
Thank you for your feedback!
– Ryan
Jul 27 at 3:02
Thank you for your feedback!
– Ryan
Jul 27 at 3:02
1
1
@Ryan You're welcome!
– Ovi
Jul 27 at 3:03
@Ryan You're welcome!
– Ovi
Jul 27 at 3:03
add a comment |Â
up vote
0
down vote
Claim: $n^2geq n$ for all $ninmathbbZ$.
If there is even a single counterexample, then we know that the claim is false. So in fact, the true negation of the claim is:
Suppose $n^2<n$ for some $ninmathbbZ$.
The rest of the proof follows like this:
If $n^2<n$ for some $ninmathbbZ$, then $n^2-n<0implies n(n-1)<0$.
Consider the following three cases:
$1)$ If $n=0$, then $n(n-1)=0nless0$ so we arrive at a contradiction.
$2)$ If $n>0$, then $n-1geq 0$ so $n(n-1)nless 0$ and we arrive at a contradiction.
$3)$ If $n<0$, then $n-1<0$ so $n(n-1)>0$ and we arrive at a contradiction.
QED
answered Jul 27 at 2:58
高田航
1,116318
1
Thank you for breaking it down so easily for me to understand.
– Ryan
Jul 27 at 3:02
add a comment |Â
up vote
0
down vote
Claim: $n^2geq n$ for all $ninmathbbZ$.
If there is even a single counterexample, then we know that the claim is false. So in fact, the true negation of the claim is:
Suppose $n^2<n$ for some $ninmathbbZ$.
The rest of the proof follows like this:
If $n^2<n$ for some $ninmathbbZ$, then $n^2-n<0implies n(n-1)<0$.
Consider the following three cases:
$1)$ If $n=0$, then $n(n-1)=0nless0$ so we arrive at a contradiction.
$2)$ If $n>0$, then $n-1geq 0$ so $n(n-1)nless 0$ and we arrive at a contradiction.
$3)$ If $n<0$, then $n-1<0$ so $n(n-1)>0$ and we arrive at a contradiction.
QED
answered Jul 27 at 2:58
高田航
1,116318
1
Thank you for breaking it down so easily for me to understand.
– Ryan
Jul 27 at 3:02
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Claim: $n^2geq n$ for all $ninmathbbZ$.
If there is even a single counterexample, then we know that the claim is false. So in fact, the true negation of the claim is:
Suppose $n^2<n$ for some $ninmathbbZ$.
The rest of the proof follows like this:
If $n^2<n$ for some $ninmathbbZ$, then $n^2-n<0implies n(n-1)<0$.
Consider the following three cases:
$1)$ If $n=0$, then $n(n-1)=0nless0$ so we arrive at a contradiction.
$2)$ If $n>0$, then $n-1geq 0$ so $n(n-1)nless 0$ and we arrive at a contradiction.
$3)$ If $n<0$, then $n-1<0$ so $n(n-1)>0$ and we arrive at a contradiction.
QED
answered Jul 27 at 2:58
高田航
1,116318
Claim: $n^2geq n$ for all $ninmathbbZ$.
If there is even a single counterexample, then we know that the claim is false. So in fact, the true negation of the claim is:
Suppose $n^2<n$ for some $ninmathbbZ$.
The rest of the proof follows like this:
If $n^2<n$ for some $ninmathbbZ$, then $n^2-n<0implies n(n-1)<0$.
Consider the following three cases:
$1)$ If $n=0$, then $n(n-1)=0nless0$ so we arrive at a contradiction.
$2)$ If $n>0$, then $n-1geq 0$ so $n(n-1)nless 0$ and we arrive at a contradiction.
$3)$ If $n<0$, then $n-1<0$ so $n(n-1)>0$ and we arrive at a contradiction.
QED
answered Jul 27 at 2:58
高田航
1,116318
answered Jul 27 at 2:58
高田航
1,116318
answered Jul 27 at 2:58
高田航
1,116318
answered Jul 27 at 2:58
高田航
1,116318
1,116318
1
Thank you for breaking it down so easily for me to understand.
– Ryan
Jul 27 at 3:02
add a comment |Â
1
Thank you for breaking it down so easily for me to understand.
– Ryan
Jul 27 at 3:02
1
1
Thank you for breaking it down so easily for me to understand.
– Ryan
Jul 27 at 3:02
Thank you for breaking it down so easily for me to understand.
– Ryan
Jul 27 at 3:02
add a comment |Â
up vote
0
down vote
The negation of "for all" is "there exists". If you want to proceed solving this question with a proof by contradiction, your starting assumption must be "Suppose $n^2<n$ for some integer $nin mathbbZ$.
Then, your need to be a little more careful in the writing of the proof. Indeed, you are trying to divide your inequality by $n$ to deduce a contradiction, but there are two things you need to be attentive to. First, $n$ must not be zero. The case $n=0$, which is trivial, must be mentioned separatedly. Second, if $n$ is negative, then the order of the inequality changes. So you need to treat the cases $n$ positive and $n$ negative separatedly (and conclude as you did).
Now, because of these cases that need to be distinguished, this kind of proof is not recommended.
You may show the result directly, by noticing that $n^2 geq n$ if and only if $n(n-1)geq 0$. A product of two quantities is non-negative if and only if both have the same sign. Thus, you are reduced to proving that for every $n in mathbbZ$, $n$ and $n-1$ have the same sign (with the meaning "both are non-negative or both are non-positive"). This is now rather straightforward.
answered Jul 27 at 2:56
Suzet
2,203427
Gotcha! Thank you so much for your help.
– Ryan
Jul 27 at 2:57
add a comment |Â
up vote
0
down vote
The negation of "for all" is "there exists". If you want to proceed solving this question with a proof by contradiction, your starting assumption must be "Suppose $n^2<n$ for some integer $nin mathbbZ$.
Then, your need to be a little more careful in the writing of the proof. Indeed, you are trying to divide your inequality by $n$ to deduce a contradiction, but there are two things you need to be attentive to. First, $n$ must not be zero. The case $n=0$, which is trivial, must be mentioned separatedly. Second, if $n$ is negative, then the order of the inequality changes. So you need to treat the cases $n$ positive and $n$ negative separatedly (and conclude as you did).
Now, because of these cases that need to be distinguished, this kind of proof is not recommended.
You may show the result directly, by noticing that $n^2 geq n$ if and only if $n(n-1)geq 0$. A product of two quantities is non-negative if and only if both have the same sign. Thus, you are reduced to proving that for every $n in mathbbZ$, $n$ and $n-1$ have the same sign (with the meaning "both are non-negative or both are non-positive"). This is now rather straightforward.
answered Jul 27 at 2:56
Suzet
2,203427
Gotcha! Thank you so much for your help.
– Ryan
Jul 27 at 2:57
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The negation of "for all" is "there exists". If you want to proceed solving this question with a proof by contradiction, your starting assumption must be "Suppose $n^2<n$ for some integer $nin mathbbZ$.
Then, your need to be a little more careful in the writing of the proof. Indeed, you are trying to divide your inequality by $n$ to deduce a contradiction, but there are two things you need to be attentive to. First, $n$ must not be zero. The case $n=0$, which is trivial, must be mentioned separatedly. Second, if $n$ is negative, then the order of the inequality changes. So you need to treat the cases $n$ positive and $n$ negative separatedly (and conclude as you did).
Now, because of these cases that need to be distinguished, this kind of proof is not recommended.
You may show the result directly, by noticing that $n^2 geq n$ if and only if $n(n-1)geq 0$. A product of two quantities is non-negative if and only if both have the same sign. Thus, you are reduced to proving that for every $n in mathbbZ$, $n$ and $n-1$ have the same sign (with the meaning "both are non-negative or both are non-positive"). This is now rather straightforward.
answered Jul 27 at 2:56
Suzet
2,203427
The negation of "for all" is "there exists". If you want to proceed solving this question with a proof by contradiction, your starting assumption must be "Suppose $n^2<n$ for some integer $nin mathbbZ$.
Then, your need to be a little more careful in the writing of the proof. Indeed, you are trying to divide your inequality by $n$ to deduce a contradiction, but there are two things you need to be attentive to. First, $n$ must not be zero. The case $n=0$, which is trivial, must be mentioned separatedly. Second, if $n$ is negative, then the order of the inequality changes. So you need to treat the cases $n$ positive and $n$ negative separatedly (and conclude as you did).
Now, because of these cases that need to be distinguished, this kind of proof is not recommended.
You may show the result directly, by noticing that $n^2 geq n$ if and only if $n(n-1)geq 0$. A product of two quantities is non-negative if and only if both have the same sign. Thus, you are reduced to proving that for every $n in mathbbZ$, $n$ and $n-1$ have the same sign (with the meaning "both are non-negative or both are non-positive"). This is now rather straightforward.
answered Jul 27 at 2:56
Suzet
2,203427
edited Jul 27 at 2:58
answered Jul 27 at 2:56
Suzet
2,203427
answered Jul 27 at 2:56
Suzet
2,203427
answered Jul 27 at 2:56
Suzet
2,203427
2,203427
Gotcha! Thank you so much for your help.
– Ryan
Jul 27 at 2:57
add a comment |Â
Gotcha! Thank you so much for your help.
– Ryan
Jul 27 at 2:57
Gotcha! Thank you so much for your help.
– Ryan
Jul 27 at 2:57
Gotcha! Thank you so much for your help.
– Ryan
Jul 27 at 2:57
add a comment |Â
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I don't see the need for proof by contradiction.
– Shrey Joshi
Jul 27 at 2:47
If you do a proof by contradiction (not recommended, but that seems to be how you are starting), the negation of: 'For all integers $n$, $n^2>n$' is not 'For all integers $n$, $n^2<n$' ; but rather: 'There exists an integer $n$ such that $n^2<n$. That is, the negation of a universal statement (for all), is an existential statement (there exists)..
– paw88789
Jul 27 at 2:53
Thanks. Would a direct proof be adequate? If so, how do I go about it? Intuitively, I can look at the claim and know it is true, but writing it in proof format is the hard part.
– Ryan
Jul 27 at 2:56
1
Both direct and indirect have the same principal. If $n ge 1$ then $n*n ge n*1$ and that works whether you divide to go from $n^2 < n implies n < 1$ which is a contradiction or if you go from $n ge 1 implies n^2 ge n$. But what would you do if some pick pendant (like me) comes along and asks "Why do you say $ncdot n < n$ would imply $n < 1$? Why do you claim that? How do you know it is true?" Would you have an answer?
– fleablood
Jul 27 at 3:12