Mixing
Regarding $x < y Rightarrow x^n < y^n$ proof rigor.
$begingroup$
I came across the implication
$$x < y Rightarrow x^n < y^n$$
$$x,y>0, nin Z^+$$
in a textbook and came up with the following proof.
Proof
Since $x<y$ the following chain of inequalities holds.
$x^n<x^{n-1}y < x^{n-2}y^2 <...<y^n$
My question now relates to the “solidity” of this proof. I get the feeling that it is a bit vague in its reasoning, yet I cannot see that it isn’t correct... I’m quite new to proving things and can’t always see the difference between intuitive reasoning and rigorous reasoning.
Is my reasoning solid? If not, what is it lacking?
proof-verification inequality proof-writing real-numbers alternative-proof
asked Jan 11 at 11:00
John DoughJohn Dough
183
$endgroup$
|
show 4 more comments
$begingroup$
I came across the implication
$$x < y Rightarrow x^n < y^n$$
$$x,y>0, nin Z^+$$
in a textbook and came up with the following proof.
Proof
Since $x<y$ the following chain of inequalities holds.
$x^n<x^{n-1}y < x^{n-2}y^2 <...<y^n$
My question now relates to the “solidity” of this proof. I get the feeling that it is a bit vague in its reasoning, yet I cannot see that it isn’t correct... I’m quite new to proving things and can’t always see the difference between intuitive reasoning and rigorous reasoning.
Is my reasoning solid? If not, what is it lacking?
proof-verification inequality proof-writing real-numbers alternative-proof
asked Jan 11 at 11:00
John DoughJohn Dough
183
$endgroup$
1
$begingroup$
It looks good to me. Maybe you could mention $x<yimplies1<y/x$ to begin with, and then each step in your induction is a multiplication by $y/x$ on both sides of the old inequality.
$endgroup$
– String
Jan 11 at 11:06
3
$begingroup$
In a word? Induction. It's perfect for lending some formality to arguments that come with "$ldots$" somewhere in them.
$endgroup$
– Theo Bendit
Jan 11 at 11:08
1
$begingroup$
Each of the inequities can be broken down to "If $a>0$ and $b<c$, then $ab<ac$." Now, I'm not sure how much rigor you're looking for. Do you want to prove the fact above via axioms for the real numbers? Or....
$endgroup$
– Michael Burr
Jan 11 at 11:08
1
$begingroup$
That seems fine to me.
$endgroup$
– Matt Samuel
Jan 11 at 15:36
1
$begingroup$
You do want to multiply both sides by the quantity though.
$endgroup$
– Matt Samuel
Jan 11 at 16:36
|
show 4 more comments
$begingroup$
I came across the implication
$$x < y Rightarrow x^n < y^n$$
$$x,y>0, nin Z^+$$
in a textbook and came up with the following proof.
Proof
Since $x<y$ the following chain of inequalities holds.
$x^n<x^{n-1}y < x^{n-2}y^2 <...<y^n$
My question now relates to the “solidity” of this proof. I get the feeling that it is a bit vague in its reasoning, yet I cannot see that it isn’t correct... I’m quite new to proving things and can’t always see the difference between intuitive reasoning and rigorous reasoning.
Is my reasoning solid? If not, what is it lacking?
proof-verification inequality proof-writing real-numbers alternative-proof
asked Jan 11 at 11:00
John DoughJohn Dough
183
$endgroup$
I came across the implication
$$x < y Rightarrow x^n < y^n$$
$$x,y>0, nin Z^+$$
in a textbook and came up with the following proof.
Proof
Since $x<y$ the following chain of inequalities holds.
$x^n<x^{n-1}y < x^{n-2}y^2 <...<y^n$
My question now relates to the “solidity” of this proof. I get the feeling that it is a bit vague in its reasoning, yet I cannot see that it isn’t correct... I’m quite new to proving things and can’t always see the difference between intuitive reasoning and rigorous reasoning.
Is my reasoning solid? If not, what is it lacking?
proof-verification inequality proof-writing real-numbers alternative-proof
proof-verification inequality proof-writing real-numbers alternative-proof
asked Jan 11 at 11:00
John DoughJohn Dough
183
asked Jan 11 at 11:00
John DoughJohn Dough
183
asked Jan 11 at 11:00
John DoughJohn Dough
183
asked Jan 11 at 11:00
John DoughJohn Dough
183
asked Jan 11 at 11:00
John DoughJohn Dough
183
183
1
$begingroup$
It looks good to me. Maybe you could mention $x<yimplies1<y/x$ to begin with, and then each step in your induction is a multiplication by $y/x$ on both sides of the old inequality.
$endgroup$
– String
Jan 11 at 11:06
3
$begingroup$
In a word? Induction. It's perfect for lending some formality to arguments that come with "$ldots$" somewhere in them.
$endgroup$
– Theo Bendit
Jan 11 at 11:08
1
$begingroup$
Each of the inequities can be broken down to "If $a>0$ and $b<c$, then $ab<ac$." Now, I'm not sure how much rigor you're looking for. Do you want to prove the fact above via axioms for the real numbers? Or....
$endgroup$
– Michael Burr
Jan 11 at 11:08
1
$begingroup$
That seems fine to me.
$endgroup$
– Matt Samuel
Jan 11 at 15:36
1
$begingroup$
You do want to multiply both sides by the quantity though.
$endgroup$
– Matt Samuel
Jan 11 at 16:36
|
show 4 more comments
1
$begingroup$
It looks good to me. Maybe you could mention $x<yimplies1<y/x$ to begin with, and then each step in your induction is a multiplication by $y/x$ on both sides of the old inequality.
$endgroup$
– String
Jan 11 at 11:06
3
$begingroup$
In a word? Induction. It's perfect for lending some formality to arguments that come with "$ldots$" somewhere in them.
$endgroup$
– Theo Bendit
Jan 11 at 11:08
1
$begingroup$
Each of the inequities can be broken down to "If $a>0$ and $b<c$, then $ab<ac$." Now, I'm not sure how much rigor you're looking for. Do you want to prove the fact above via axioms for the real numbers? Or....
$endgroup$
– Michael Burr
Jan 11 at 11:08
1
$begingroup$
That seems fine to me.
$endgroup$
– Matt Samuel
Jan 11 at 15:36
1
$begingroup$
You do want to multiply both sides by the quantity though.
$endgroup$
– Matt Samuel
Jan 11 at 16:36
1
1
$begingroup$
It looks good to me. Maybe you could mention $x<yimplies1<y/x$ to begin with, and then each step in your induction is a multiplication by $y/x$ on both sides of the old inequality.
$endgroup$
– String
Jan 11 at 11:06
$begingroup$
It looks good to me. Maybe you could mention $x<yimplies1<y/x$ to begin with, and then each step in your induction is a multiplication by $y/x$ on both sides of the old inequality.
$endgroup$
– String
Jan 11 at 11:06
3
3
$begingroup$
In a word? Induction. It's perfect for lending some formality to arguments that come with "$ldots$" somewhere in them.
$endgroup$
– Theo Bendit
Jan 11 at 11:08
$begingroup$
In a word? Induction. It's perfect for lending some formality to arguments that come with "$ldots$" somewhere in them.
$endgroup$
– Theo Bendit
Jan 11 at 11:08
1
1
$begingroup$
Each of the inequities can be broken down to "If $a>0$ and $b<c$, then $ab<ac$." Now, I'm not sure how much rigor you're looking for. Do you want to prove the fact above via axioms for the real numbers? Or....
$endgroup$
– Michael Burr
Jan 11 at 11:08
$begingroup$
Each of the inequities can be broken down to "If $a>0$ and $b<c$, then $ab<ac$." Now, I'm not sure how much rigor you're looking for. Do you want to prove the fact above via axioms for the real numbers? Or....
$endgroup$
– Michael Burr
Jan 11 at 11:08
1
1
$begingroup$
That seems fine to me.
$endgroup$
– Matt Samuel
Jan 11 at 15:36
$begingroup$
That seems fine to me.
$endgroup$
– Matt Samuel
Jan 11 at 15:36
1
1
$begingroup$
You do want to multiply both sides by the quantity though.
$endgroup$
– Matt Samuel
Jan 11 at 16:36
$begingroup$
You do want to multiply both sides by the quantity though.
$endgroup$
– Matt Samuel
Jan 11 at 16:36
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
What's lacking is explicit induction. It's an "et cetera" argument. It's not wrong, but it's not $100%$ rigorous. It's close enough that making it rigorous is easy though, if you know how to do induction.
answered Jan 11 at 11:07
Matt SamuelMatt Samuel
38.1k63767
$endgroup$
$begingroup$
Thank you for this answer! I was slightly perplexed about how to apply induction to my proof but think I figured it out in my comment to the question. Does it seem correct?
$endgroup$
– John Dough
Jan 11 at 15:16
add a comment |
$begingroup$
Ask yourself the following question: Where in the proof do I (tacitly) use the hypothesis that $x$ and $y$ are positive? Then make that use explicit.
It might also help to rewrite the proof as a proof by induction, inducting on $n$. In other words, show that if $0lt xlt y$ and $x^nlt y^n$, then $x^{n+1}lt y^{n+1}$ (again, making explicit where the positivity hypothesis is used).
Remark: In general, it's a good idea, when either writing or reading a proof, to ask, what are the hypotheses and where in the proof are they used? It sometimes happens that a proof does not use a hypothesis, even tacitly, which sometimes means that the hypothesis was unnecessary -- and sometimes means the proof is wrong!
answered Jan 11 at 11:24
Barry CipraBarry Cipra
59.6k653126
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
What's lacking is explicit induction. It's an "et cetera" argument. It's not wrong, but it's not $100%$ rigorous. It's close enough that making it rigorous is easy though, if you know how to do induction.
answered Jan 11 at 11:07
Matt SamuelMatt Samuel
38.1k63767
$endgroup$
$begingroup$
Thank you for this answer! I was slightly perplexed about how to apply induction to my proof but think I figured it out in my comment to the question. Does it seem correct?
$endgroup$
– John Dough
Jan 11 at 15:16
add a comment |
$begingroup$
What's lacking is explicit induction. It's an "et cetera" argument. It's not wrong, but it's not $100%$ rigorous. It's close enough that making it rigorous is easy though, if you know how to do induction.
answered Jan 11 at 11:07
Matt SamuelMatt Samuel
38.1k63767
$endgroup$
$begingroup$
Thank you for this answer! I was slightly perplexed about how to apply induction to my proof but think I figured it out in my comment to the question. Does it seem correct?
$endgroup$
– John Dough
Jan 11 at 15:16
add a comment |
$begingroup$
What's lacking is explicit induction. It's an "et cetera" argument. It's not wrong, but it's not $100%$ rigorous. It's close enough that making it rigorous is easy though, if you know how to do induction.
answered Jan 11 at 11:07
Matt SamuelMatt Samuel
38.1k63767
$endgroup$
What's lacking is explicit induction. It's an "et cetera" argument. It's not wrong, but it's not $100%$ rigorous. It's close enough that making it rigorous is easy though, if you know how to do induction.
answered Jan 11 at 11:07
Matt SamuelMatt Samuel
38.1k63767
answered Jan 11 at 11:07
Matt SamuelMatt Samuel
38.1k63767
answered Jan 11 at 11:07
Matt SamuelMatt Samuel
38.1k63767
answered Jan 11 at 11:07
Matt SamuelMatt Samuel
38.1k63767
38.1k63767
$begingroup$
Thank you for this answer! I was slightly perplexed about how to apply induction to my proof but think I figured it out in my comment to the question. Does it seem correct?
$endgroup$
– John Dough
Jan 11 at 15:16
add a comment |
$begingroup$
Thank you for this answer! I was slightly perplexed about how to apply induction to my proof but think I figured it out in my comment to the question. Does it seem correct?
$endgroup$
– John Dough
Jan 11 at 15:16
$begingroup$
Thank you for this answer! I was slightly perplexed about how to apply induction to my proof but think I figured it out in my comment to the question. Does it seem correct?
$endgroup$
– John Dough
Jan 11 at 15:16
$begingroup$
Thank you for this answer! I was slightly perplexed about how to apply induction to my proof but think I figured it out in my comment to the question. Does it seem correct?
$endgroup$
– John Dough
Jan 11 at 15:16
add a comment |
$begingroup$
Ask yourself the following question: Where in the proof do I (tacitly) use the hypothesis that $x$ and $y$ are positive? Then make that use explicit.
It might also help to rewrite the proof as a proof by induction, inducting on $n$. In other words, show that if $0lt xlt y$ and $x^nlt y^n$, then $x^{n+1}lt y^{n+1}$ (again, making explicit where the positivity hypothesis is used).
Remark: In general, it's a good idea, when either writing or reading a proof, to ask, what are the hypotheses and where in the proof are they used? It sometimes happens that a proof does not use a hypothesis, even tacitly, which sometimes means that the hypothesis was unnecessary -- and sometimes means the proof is wrong!
answered Jan 11 at 11:24
Barry CipraBarry Cipra
59.6k653126
$endgroup$
add a comment |
$begingroup$
Ask yourself the following question: Where in the proof do I (tacitly) use the hypothesis that $x$ and $y$ are positive? Then make that use explicit.
It might also help to rewrite the proof as a proof by induction, inducting on $n$. In other words, show that if $0lt xlt y$ and $x^nlt y^n$, then $x^{n+1}lt y^{n+1}$ (again, making explicit where the positivity hypothesis is used).
Remark: In general, it's a good idea, when either writing or reading a proof, to ask, what are the hypotheses and where in the proof are they used? It sometimes happens that a proof does not use a hypothesis, even tacitly, which sometimes means that the hypothesis was unnecessary -- and sometimes means the proof is wrong!
answered Jan 11 at 11:24
Barry CipraBarry Cipra
59.6k653126
$endgroup$
add a comment |
$begingroup$
Ask yourself the following question: Where in the proof do I (tacitly) use the hypothesis that $x$ and $y$ are positive? Then make that use explicit.
It might also help to rewrite the proof as a proof by induction, inducting on $n$. In other words, show that if $0lt xlt y$ and $x^nlt y^n$, then $x^{n+1}lt y^{n+1}$ (again, making explicit where the positivity hypothesis is used).
Remark: In general, it's a good idea, when either writing or reading a proof, to ask, what are the hypotheses and where in the proof are they used? It sometimes happens that a proof does not use a hypothesis, even tacitly, which sometimes means that the hypothesis was unnecessary -- and sometimes means the proof is wrong!
answered Jan 11 at 11:24
Barry CipraBarry Cipra
59.6k653126
$endgroup$
Ask yourself the following question: Where in the proof do I (tacitly) use the hypothesis that $x$ and $y$ are positive? Then make that use explicit.
It might also help to rewrite the proof as a proof by induction, inducting on $n$. In other words, show that if $0lt xlt y$ and $x^nlt y^n$, then $x^{n+1}lt y^{n+1}$ (again, making explicit where the positivity hypothesis is used).
Remark: In general, it's a good idea, when either writing or reading a proof, to ask, what are the hypotheses and where in the proof are they used? It sometimes happens that a proof does not use a hypothesis, even tacitly, which sometimes means that the hypothesis was unnecessary -- and sometimes means the proof is wrong!
answered Jan 11 at 11:24
Barry CipraBarry Cipra
59.6k653126
edited Jan 11 at 11:31
answered Jan 11 at 11:24
Barry CipraBarry Cipra
59.6k653126
answered Jan 11 at 11:24
Barry CipraBarry Cipra
59.6k653126
answered Jan 11 at 11:24
Barry CipraBarry Cipra
59.6k653126
59.6k653126
add a comment |
add a comment |
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1
$begingroup$
It looks good to me. Maybe you could mention $x<yimplies1<y/x$ to begin with, and then each step in your induction is a multiplication by $y/x$ on both sides of the old inequality.
$endgroup$
– String
Jan 11 at 11:06
3
$begingroup$
In a word? Induction. It's perfect for lending some formality to arguments that come with "$ldots$" somewhere in them.
$endgroup$
– Theo Bendit
Jan 11 at 11:08
1
$begingroup$
Each of the inequities can be broken down to "If $a>0$ and $b<c$, then $ab<ac$." Now, I'm not sure how much rigor you're looking for. Do you want to prove the fact above via axioms for the real numbers? Or....
$endgroup$
– Michael Burr
Jan 11 at 11:08
1
$begingroup$
That seems fine to me.
$endgroup$
– Matt Samuel
Jan 11 at 15:36
1
$begingroup$
You do want to multiply both sides by the quantity though.
$endgroup$
– Matt Samuel
Jan 11 at 16:36