Proving that between any two real numbers there exist a real number












2














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 $x lt z lt y$.



Is this equivalent to proving the real line is continuous?






real-numbers







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  • 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?
    – user247327
    1 hour ago








  • 1




    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.
    – ÍgjøgnumMeg
    1 hour ago
















2














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 $x lt z lt y$.



Is this equivalent to proving the real line is continuous?






real-numbers







share|cite|improve this question






















  • 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?
    – user247327
    1 hour ago








  • 1




    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.
    – ÍgjøgnumMeg
    1 hour ago














2












2








2







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 $x lt z lt y$.



Is this equivalent to proving the real line is continuous?






real-numbers







share|cite|improve this question













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 $x lt z lt y$.



Is this equivalent to proving the real line is continuous?






real-numbers




real-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour ago









MHall

757




757












  • 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?
    – user247327
    1 hour ago








  • 1




    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.
    – ÍgjøgnumMeg
    1 hour ago


















  • 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?
    – user247327
    1 hour ago








  • 1




    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.
    – ÍgjøgnumMeg
    1 hour ago
















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?
– user247327
1 hour ago






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?
– user247327
1 hour ago






1




1




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.
– ÍgjøgnumMeg
1 hour ago




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.
– ÍgjøgnumMeg
1 hour ago










1 Answer
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10














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?







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    1 Answer
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    10














    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?







    share|cite|improve this answer


























      10














      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?







      share|cite|improve this answer
























        10












        10








        10






        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?







        share|cite|improve this answer












        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?








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        Noah Schweber

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