Proving that between any two real numbers there exist a real number [closed]
$begingroup$
Formally, I want to prove that if $x$ and $y$ are real numbers such that $x lt y$ then there exists a real number $z$ such that $x lt z lt y$.
I want to know whether, in constructing the proof, I should think in terms of continuity of the real line or not. In other words, is the proof of the existence of a real between any two reals equivalent to proving the real line is continuous?
This is a question in a calculus textbook I am going through as part of a self study project to brush up on my calculus. The text is called Calculus Volume 1 written by Tom M. Apostol page 19 exercise 1.14*.
real-numbers
asked Dec 28 '18 at 21:48
MHallMHall
889
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closed as off-topic by Holo, José Carlos Santos, Saad, RRL, TheSimpliFire Dec 29 '18 at 16:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Holo, José Carlos Santos, Saad, RRL, TheSimpliFire
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Formally, I want to prove that if $x$ and $y$ are real numbers such that $x lt y$ then there exists a real number $z$ such that $x lt z lt y$.
I want to know whether, in constructing the proof, I should think in terms of continuity of the real line or not. In other words, is the proof of the existence of a real between any two reals equivalent to proving the real line is continuous?
This is a question in a calculus textbook I am going through as part of a self study project to brush up on my calculus. The text is called Calculus Volume 1 written by Tom M. Apostol page 19 exercise 1.14*.
real-numbers
asked Dec 28 '18 at 21:48
MHallMHall
889
$endgroup$
closed as off-topic by Holo, José Carlos Santos, Saad, RRL, TheSimpliFire Dec 29 '18 at 16:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Holo, José Carlos Santos, Saad, RRL, TheSimpliFire
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Can you prove that "if x and y are real numbers then x+ y is a real number" (that the real numbers are closed under addition)? Can you prove then that (x+ y)/2 is a real number? Do you see that "if x< y then 2x< x+ y? And then that x< (x+ y)? What if y< x?
$endgroup$
– user247327
Dec 28 '18 at 21:52
4
$begingroup$
By "the real line is continuous" I assume you mean something (informally) along the lines of "there are no gaps", in which case, this is because $Bbb R$ is a complete metric space.
$endgroup$
– ÍgjøgnumMeg
Dec 28 '18 at 21:53
$begingroup$
i would ask, what is the definition of "less than"? is it defined by subtraction? if so, then how do you define subtraction of two real numbers? if i give you any two real numbers, how can you subtract them and what is the result? as someone who leans towards finitism, i must confess i believe this is equivalent to asking if there is a number between two infinite numbers, but i am in a tiny minority. the interesting thing to me is that the operation of subtraction has a different relationship to the infinites than it does to the finites, in terms of output type and input type.
$endgroup$
– don bright
Dec 29 '18 at 15:35
add a comment |
$begingroup$
Formally, I want to prove that if $x$ and $y$ are real numbers such that $x lt y$ then there exists a real number $z$ such that $x lt z lt y$.
I want to know whether, in constructing the proof, I should think in terms of continuity of the real line or not. In other words, is the proof of the existence of a real between any two reals equivalent to proving the real line is continuous?
This is a question in a calculus textbook I am going through as part of a self study project to brush up on my calculus. The text is called Calculus Volume 1 written by Tom M. Apostol page 19 exercise 1.14*.
real-numbers
asked Dec 28 '18 at 21:48
MHallMHall
889
$endgroup$
Formally, I want to prove that if $x$ and $y$ are real numbers such that $x lt y$ then there exists a real number $z$ such that $x lt z lt y$.
I want to know whether, in constructing the proof, I should think in terms of continuity of the real line or not. In other words, is the proof of the existence of a real between any two reals equivalent to proving the real line is continuous?
This is a question in a calculus textbook I am going through as part of a self study project to brush up on my calculus. The text is called Calculus Volume 1 written by Tom M. Apostol page 19 exercise 1.14*.
real-numbers
real-numbers
asked Dec 28 '18 at 21:48
MHallMHall
889
asked Dec 28 '18 at 21:48
MHallMHall
889
edited Dec 30 '18 at 19:13
MHall
asked Dec 28 '18 at 21:48
MHallMHall
889
asked Dec 28 '18 at 21:48
MHallMHall
889
asked Dec 28 '18 at 21:48
MHallMHall
889
889
closed as off-topic by Holo, José Carlos Santos, Saad, RRL, TheSimpliFire Dec 29 '18 at 16:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Holo, José Carlos Santos, Saad, RRL, TheSimpliFire
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Holo, José Carlos Santos, Saad, RRL, TheSimpliFire Dec 29 '18 at 16:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Holo, José Carlos Santos, Saad, RRL, TheSimpliFire
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Can you prove that "if x and y are real numbers then x+ y is a real number" (that the real numbers are closed under addition)? Can you prove then that (x+ y)/2 is a real number? Do you see that "if x< y then 2x< x+ y? And then that x< (x+ y)? What if y< x?
$endgroup$
– user247327
Dec 28 '18 at 21:52
4
$begingroup$
By "the real line is continuous" I assume you mean something (informally) along the lines of "there are no gaps", in which case, this is because $Bbb R$ is a complete metric space.
$endgroup$
– ÍgjøgnumMeg
Dec 28 '18 at 21:53
$begingroup$
i would ask, what is the definition of "less than"? is it defined by subtraction? if so, then how do you define subtraction of two real numbers? if i give you any two real numbers, how can you subtract them and what is the result? as someone who leans towards finitism, i must confess i believe this is equivalent to asking if there is a number between two infinite numbers, but i am in a tiny minority. the interesting thing to me is that the operation of subtraction has a different relationship to the infinites than it does to the finites, in terms of output type and input type.
$endgroup$
– don bright
Dec 29 '18 at 15:35
add a comment |
$begingroup$
Can you prove that "if x and y are real numbers then x+ y is a real number" (that the real numbers are closed under addition)? Can you prove then that (x+ y)/2 is a real number? Do you see that "if x< y then 2x< x+ y? And then that x< (x+ y)? What if y< x?
$endgroup$
– user247327
Dec 28 '18 at 21:52
4
$begingroup$
By "the real line is continuous" I assume you mean something (informally) along the lines of "there are no gaps", in which case, this is because $Bbb R$ is a complete metric space.
$endgroup$
– ÍgjøgnumMeg
Dec 28 '18 at 21:53
$begingroup$
i would ask, what is the definition of "less than"? is it defined by subtraction? if so, then how do you define subtraction of two real numbers? if i give you any two real numbers, how can you subtract them and what is the result? as someone who leans towards finitism, i must confess i believe this is equivalent to asking if there is a number between two infinite numbers, but i am in a tiny minority. the interesting thing to me is that the operation of subtraction has a different relationship to the infinites than it does to the finites, in terms of output type and input type.
$endgroup$
– don bright
Dec 29 '18 at 15:35
$begingroup$
Can you prove that "if x and y are real numbers then x+ y is a real number" (that the real numbers are closed under addition)? Can you prove then that (x+ y)/2 is a real number? Do you see that "if x< y then 2x< x+ y? And then that x< (x+ y)? What if y< x?
$endgroup$
– user247327
Dec 28 '18 at 21:52
$begingroup$
Can you prove that "if x and y are real numbers then x+ y is a real number" (that the real numbers are closed under addition)? Can you prove then that (x+ y)/2 is a real number? Do you see that "if x< y then 2x< x+ y? And then that x< (x+ y)? What if y< x?
$endgroup$
– user247327
Dec 28 '18 at 21:52
4
4
$begingroup$
By "the real line is continuous" I assume you mean something (informally) along the lines of "there are no gaps", in which case, this is because $Bbb R$ is a complete metric space.
$endgroup$
– ÍgjøgnumMeg
Dec 28 '18 at 21:53
$begingroup$
By "the real line is continuous" I assume you mean something (informally) along the lines of "there are no gaps", in which case, this is because $Bbb R$ is a complete metric space.
$endgroup$
– ÍgjøgnumMeg
Dec 28 '18 at 21:53
$begingroup$
i would ask, what is the definition of "less than"? is it defined by subtraction? if so, then how do you define subtraction of two real numbers? if i give you any two real numbers, how can you subtract them and what is the result? as someone who leans towards finitism, i must confess i believe this is equivalent to asking if there is a number between two infinite numbers, but i am in a tiny minority. the interesting thing to me is that the operation of subtraction has a different relationship to the infinites than it does to the finites, in terms of output type and input type.
$endgroup$
– don bright
Dec 29 '18 at 15:35
$begingroup$
i would ask, what is the definition of "less than"? is it defined by subtraction? if so, then how do you define subtraction of two real numbers? if i give you any two real numbers, how can you subtract them and what is the result? as someone who leans towards finitism, i must confess i believe this is equivalent to asking if there is a number between two infinite numbers, but i am in a tiny minority. the interesting thing to me is that the operation of subtraction has a different relationship to the infinites than it does to the finites, in terms of output type and input type.
$endgroup$
– don bright
Dec 29 '18 at 15:35
add a comment |
1 Answer
1
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oldest
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$begingroup$
First, let me say a bit about the end of your question. I'm not sure what you mean by "the real line is continuous," but I suspect you mean that the real line is connected or complete. If so, then no, that's a different thing; for example, $mathbb{Q}$ has the "density" property here (between any two rationals, there's another rational) but is neither connected (consider $(-infty,pi)$ versus $(pi,infty)$, for example) or complete (consider the sequence $3,3.1,3.14,3.141,...$, for example).
Since connectedness/completeness can be a bit technical at first, let me say the following: intuitively, a space is connected if it doesn't "break into a bunch of separate pieces," and is complete if it "doesn't have any gaps" - the set $mathbb{Q}$ of rationals, even though it's dense, has lots of gaps and breaks apart really easily.
As to proving the claim you're looking at, I suspect you're thinking too hard about it. First, try to guess what a good formula for such a $z$ might be, and then try to prove (using whatever formal system you've been given) that it actually works. And this formula will be quite simple.
HINT: if on a quiz Sam scored $10/10$ and Alex scored $6/10$, then together they got an ---- score of $8/10$. Now, how do you calculate the "---" of two numbers?
answered Dec 28 '18 at 21:52
Noah SchweberNoah Schweber
129k10152294
$endgroup$
$begingroup$
Thanks! I definitely was overthinking the problem.
$endgroup$
– MHall
Jan 1 at 18:54
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First, let me say a bit about the end of your question. I'm not sure what you mean by "the real line is continuous," but I suspect you mean that the real line is connected or complete. If so, then no, that's a different thing; for example, $mathbb{Q}$ has the "density" property here (between any two rationals, there's another rational) but is neither connected (consider $(-infty,pi)$ versus $(pi,infty)$, for example) or complete (consider the sequence $3,3.1,3.14,3.141,...$, for example).
Since connectedness/completeness can be a bit technical at first, let me say the following: intuitively, a space is connected if it doesn't "break into a bunch of separate pieces," and is complete if it "doesn't have any gaps" - the set $mathbb{Q}$ of rationals, even though it's dense, has lots of gaps and breaks apart really easily.
As to proving the claim you're looking at, I suspect you're thinking too hard about it. First, try to guess what a good formula for such a $z$ might be, and then try to prove (using whatever formal system you've been given) that it actually works. And this formula will be quite simple.
HINT: if on a quiz Sam scored $10/10$ and Alex scored $6/10$, then together they got an ---- score of $8/10$. Now, how do you calculate the "---" of two numbers?
answered Dec 28 '18 at 21:52
Noah SchweberNoah Schweber
129k10152294
$endgroup$
$begingroup$
Thanks! I definitely was overthinking the problem.
$endgroup$
– MHall
Jan 1 at 18:54
add a comment |
$begingroup$
First, let me say a bit about the end of your question. I'm not sure what you mean by "the real line is continuous," but I suspect you mean that the real line is connected or complete. If so, then no, that's a different thing; for example, $mathbb{Q}$ has the "density" property here (between any two rationals, there's another rational) but is neither connected (consider $(-infty,pi)$ versus $(pi,infty)$, for example) or complete (consider the sequence $3,3.1,3.14,3.141,...$, for example).
Since connectedness/completeness can be a bit technical at first, let me say the following: intuitively, a space is connected if it doesn't "break into a bunch of separate pieces," and is complete if it "doesn't have any gaps" - the set $mathbb{Q}$ of rationals, even though it's dense, has lots of gaps and breaks apart really easily.
As to proving the claim you're looking at, I suspect you're thinking too hard about it. First, try to guess what a good formula for such a $z$ might be, and then try to prove (using whatever formal system you've been given) that it actually works. And this formula will be quite simple.
HINT: if on a quiz Sam scored $10/10$ and Alex scored $6/10$, then together they got an ---- score of $8/10$. Now, how do you calculate the "---" of two numbers?
answered Dec 28 '18 at 21:52
Noah SchweberNoah Schweber
129k10152294
$endgroup$
$begingroup$
Thanks! I definitely was overthinking the problem.
$endgroup$
– MHall
Jan 1 at 18:54
add a comment |
$begingroup$
First, let me say a bit about the end of your question. I'm not sure what you mean by "the real line is continuous," but I suspect you mean that the real line is connected or complete. If so, then no, that's a different thing; for example, $mathbb{Q}$ has the "density" property here (between any two rationals, there's another rational) but is neither connected (consider $(-infty,pi)$ versus $(pi,infty)$, for example) or complete (consider the sequence $3,3.1,3.14,3.141,...$, for example).
Since connectedness/completeness can be a bit technical at first, let me say the following: intuitively, a space is connected if it doesn't "break into a bunch of separate pieces," and is complete if it "doesn't have any gaps" - the set $mathbb{Q}$ of rationals, even though it's dense, has lots of gaps and breaks apart really easily.
As to proving the claim you're looking at, I suspect you're thinking too hard about it. First, try to guess what a good formula for such a $z$ might be, and then try to prove (using whatever formal system you've been given) that it actually works. And this formula will be quite simple.
HINT: if on a quiz Sam scored $10/10$ and Alex scored $6/10$, then together they got an ---- score of $8/10$. Now, how do you calculate the "---" of two numbers?
answered Dec 28 '18 at 21:52
Noah SchweberNoah Schweber
129k10152294
$endgroup$
First, let me say a bit about the end of your question. I'm not sure what you mean by "the real line is continuous," but I suspect you mean that the real line is connected or complete. If so, then no, that's a different thing; for example, $mathbb{Q}$ has the "density" property here (between any two rationals, there's another rational) but is neither connected (consider $(-infty,pi)$ versus $(pi,infty)$, for example) or complete (consider the sequence $3,3.1,3.14,3.141,...$, for example).
Since connectedness/completeness can be a bit technical at first, let me say the following: intuitively, a space is connected if it doesn't "break into a bunch of separate pieces," and is complete if it "doesn't have any gaps" - the set $mathbb{Q}$ of rationals, even though it's dense, has lots of gaps and breaks apart really easily.
As to proving the claim you're looking at, I suspect you're thinking too hard about it. First, try to guess what a good formula for such a $z$ might be, and then try to prove (using whatever formal system you've been given) that it actually works. And this formula will be quite simple.
HINT: if on a quiz Sam scored $10/10$ and Alex scored $6/10$, then together they got an ---- score of $8/10$. Now, how do you calculate the "---" of two numbers?
answered Dec 28 '18 at 21:52
Noah SchweberNoah Schweber
129k10152294
answered Dec 28 '18 at 21:52
Noah SchweberNoah Schweber
129k10152294
answered Dec 28 '18 at 21:52
Noah SchweberNoah Schweber
129k10152294
answered Dec 28 '18 at 21:52
Noah SchweberNoah Schweber
129k10152294
129k10152294
$begingroup$
Thanks! I definitely was overthinking the problem.
$endgroup$
– MHall
Jan 1 at 18:54
add a comment |
$begingroup$
Thanks! I definitely was overthinking the problem.
$endgroup$
– MHall
Jan 1 at 18:54
$begingroup$
Thanks! I definitely was overthinking the problem.
$endgroup$
– MHall
Jan 1 at 18:54
$begingroup$
Thanks! I definitely was overthinking the problem.
$endgroup$
– MHall
Jan 1 at 18:54
add a comment |
$begingroup$
Can you prove that "if x and y are real numbers then x+ y is a real number" (that the real numbers are closed under addition)? Can you prove then that (x+ y)/2 is a real number? Do you see that "if x< y then 2x< x+ y? And then that x< (x+ y)? What if y< x?
$endgroup$
– user247327
Dec 28 '18 at 21:52
4
$begingroup$
By "the real line is continuous" I assume you mean something (informally) along the lines of "there are no gaps", in which case, this is because $Bbb R$ is a complete metric space.
$endgroup$
– ÍgjøgnumMeg
Dec 28 '18 at 21:53
$begingroup$
i would ask, what is the definition of "less than"? is it defined by subtraction? if so, then how do you define subtraction of two real numbers? if i give you any two real numbers, how can you subtract them and what is the result? as someone who leans towards finitism, i must confess i believe this is equivalent to asking if there is a number between two infinite numbers, but i am in a tiny minority. the interesting thing to me is that the operation of subtraction has a different relationship to the infinites than it does to the finites, in terms of output type and input type.
$endgroup$
– don bright
Dec 29 '18 at 15:35