Why there are irrational numbers?











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6
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I do not quite get it. Why can't we represent all real numbers as a sum of rational numbers? Why do we need irrational numbers?



For example,




  • $pi=3.14159265358cdots=3+10^{-1}+4*10^{-2}+10^{-3}+5*10^{-4}+cdots$

  • $e=2.71828182846cdots=2+7*10^{-1}+10^{-2}+8*10^{-3}+2*10^{-4}+cdots$

  • And so on










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  • 19




    For completeness ;)
    – Daniel Fischer
    Jun 24 '14 at 18:46






  • 9




    Those aren't sums, those are limits of sequences.
    – Git Gud
    Jun 24 '14 at 18:47






  • 7




    $frac12$ is not a sum of integers, no matter how hard you try.
    – Asaf Karagila
    Jun 24 '14 at 18:49






  • 1




    What is the last rational number in one of your sums?
    – J. W. Perry
    Jun 24 '14 at 20:11






  • 3




    That is much more clear. The answer to your question is: any finite sum of rationals is a rational, but that is not the case for the limit of an infinite sum. More generally, get into your head now that a property shared by every member of a set and a property of the limit of that set need not be the same. Every one of your finite sums is rational, but the limit need not be. Accidentally treating set elements and limits as the same thing is the cause of a great many mathematical mistakes.
    – Eric Lippert
    Jun 24 '14 at 22:15

















up vote
6
down vote

favorite












I do not quite get it. Why can't we represent all real numbers as a sum of rational numbers? Why do we need irrational numbers?



For example,




  • $pi=3.14159265358cdots=3+10^{-1}+4*10^{-2}+10^{-3}+5*10^{-4}+cdots$

  • $e=2.71828182846cdots=2+7*10^{-1}+10^{-2}+8*10^{-3}+2*10^{-4}+cdots$

  • And so on










share|cite|improve this question




















  • 19




    For completeness ;)
    – Daniel Fischer
    Jun 24 '14 at 18:46






  • 9




    Those aren't sums, those are limits of sequences.
    – Git Gud
    Jun 24 '14 at 18:47






  • 7




    $frac12$ is not a sum of integers, no matter how hard you try.
    – Asaf Karagila
    Jun 24 '14 at 18:49






  • 1




    What is the last rational number in one of your sums?
    – J. W. Perry
    Jun 24 '14 at 20:11






  • 3




    That is much more clear. The answer to your question is: any finite sum of rationals is a rational, but that is not the case for the limit of an infinite sum. More generally, get into your head now that a property shared by every member of a set and a property of the limit of that set need not be the same. Every one of your finite sums is rational, but the limit need not be. Accidentally treating set elements and limits as the same thing is the cause of a great many mathematical mistakes.
    – Eric Lippert
    Jun 24 '14 at 22:15















up vote
6
down vote

favorite









up vote
6
down vote

favorite











I do not quite get it. Why can't we represent all real numbers as a sum of rational numbers? Why do we need irrational numbers?



For example,




  • $pi=3.14159265358cdots=3+10^{-1}+4*10^{-2}+10^{-3}+5*10^{-4}+cdots$

  • $e=2.71828182846cdots=2+7*10^{-1}+10^{-2}+8*10^{-3}+2*10^{-4}+cdots$

  • And so on










share|cite|improve this question















I do not quite get it. Why can't we represent all real numbers as a sum of rational numbers? Why do we need irrational numbers?



For example,




  • $pi=3.14159265358cdots=3+10^{-1}+4*10^{-2}+10^{-3}+5*10^{-4}+cdots$

  • $e=2.71828182846cdots=2+7*10^{-1}+10^{-2}+8*10^{-3}+2*10^{-4}+cdots$

  • And so on







irrational-numbers






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edited Jun 24 '14 at 19:23









goblin

36.4k1157188




36.4k1157188










asked Jun 24 '14 at 18:45









Kira

10610




10610








  • 19




    For completeness ;)
    – Daniel Fischer
    Jun 24 '14 at 18:46






  • 9




    Those aren't sums, those are limits of sequences.
    – Git Gud
    Jun 24 '14 at 18:47






  • 7




    $frac12$ is not a sum of integers, no matter how hard you try.
    – Asaf Karagila
    Jun 24 '14 at 18:49






  • 1




    What is the last rational number in one of your sums?
    – J. W. Perry
    Jun 24 '14 at 20:11






  • 3




    That is much more clear. The answer to your question is: any finite sum of rationals is a rational, but that is not the case for the limit of an infinite sum. More generally, get into your head now that a property shared by every member of a set and a property of the limit of that set need not be the same. Every one of your finite sums is rational, but the limit need not be. Accidentally treating set elements and limits as the same thing is the cause of a great many mathematical mistakes.
    – Eric Lippert
    Jun 24 '14 at 22:15
















  • 19




    For completeness ;)
    – Daniel Fischer
    Jun 24 '14 at 18:46






  • 9




    Those aren't sums, those are limits of sequences.
    – Git Gud
    Jun 24 '14 at 18:47






  • 7




    $frac12$ is not a sum of integers, no matter how hard you try.
    – Asaf Karagila
    Jun 24 '14 at 18:49






  • 1




    What is the last rational number in one of your sums?
    – J. W. Perry
    Jun 24 '14 at 20:11






  • 3




    That is much more clear. The answer to your question is: any finite sum of rationals is a rational, but that is not the case for the limit of an infinite sum. More generally, get into your head now that a property shared by every member of a set and a property of the limit of that set need not be the same. Every one of your finite sums is rational, but the limit need not be. Accidentally treating set elements and limits as the same thing is the cause of a great many mathematical mistakes.
    – Eric Lippert
    Jun 24 '14 at 22:15










19




19




For completeness ;)
– Daniel Fischer
Jun 24 '14 at 18:46




For completeness ;)
– Daniel Fischer
Jun 24 '14 at 18:46




9




9




Those aren't sums, those are limits of sequences.
– Git Gud
Jun 24 '14 at 18:47




Those aren't sums, those are limits of sequences.
– Git Gud
Jun 24 '14 at 18:47




7




7




$frac12$ is not a sum of integers, no matter how hard you try.
– Asaf Karagila
Jun 24 '14 at 18:49




$frac12$ is not a sum of integers, no matter how hard you try.
– Asaf Karagila
Jun 24 '14 at 18:49




1




1




What is the last rational number in one of your sums?
– J. W. Perry
Jun 24 '14 at 20:11




What is the last rational number in one of your sums?
– J. W. Perry
Jun 24 '14 at 20:11




3




3




That is much more clear. The answer to your question is: any finite sum of rationals is a rational, but that is not the case for the limit of an infinite sum. More generally, get into your head now that a property shared by every member of a set and a property of the limit of that set need not be the same. Every one of your finite sums is rational, but the limit need not be. Accidentally treating set elements and limits as the same thing is the cause of a great many mathematical mistakes.
– Eric Lippert
Jun 24 '14 at 22:15






That is much more clear. The answer to your question is: any finite sum of rationals is a rational, but that is not the case for the limit of an infinite sum. More generally, get into your head now that a property shared by every member of a set and a property of the limit of that set need not be the same. Every one of your finite sums is rational, but the limit need not be. Accidentally treating set elements and limits as the same thing is the cause of a great many mathematical mistakes.
– Eric Lippert
Jun 24 '14 at 22:15












5 Answers
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active

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up vote
7
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You can represent any real number as some convergent sequence of rational numbers, as you do above. However, irrational numbers are those numbers that cannot be expressed as any finite such sequence. However long of a finite sequence of rational numbers approaching $pi$ you have, there is a positive real number $varepsilon$ such that the last term in the sequence is at least $varepsilon$ away from $pi$ (In particular, we can take the last term of the sequence, $t$, and take $varepsilon = frac{|pi - t|}{2}$).






share|cite|improve this answer




























    up vote
    5
    down vote













    For a long time people believed that the rational numbers were enough to perform any arithmetic operation we needed. It wasn't until 500 BC that the Pythagoreans started to become aware that the rational numbers were not quite sufficient.



    For instance, the Pythagoreans tried to compute something that seemed benign. They wanted to find the length of the diagonal of a square with sides of unit length. We know now (by the Pythagorean theorem) that this should be the number we represent as $sqrt{2}$. The Pythagoreans tried to find a rational number corresponding to this and came to a contradiction. It's said that the Pythagoreans were so upset that they drowned the person who discovered this. I have also heard speculation that this is why Greek mathematics was so focused on geometry, since this is something you can "see" and away from uncomfortable concepts like irrational numbers.



    This was the first instance where irrational number started to seem necessary. The next big irrational number to come forward was the number $pi$. This was in 350 BC, and the first to try to approximate this number was Archimedes. This is where the estimate 22/7 came from. He also had a better estimate than that, it escapes me at the moment. The Bible also estimates this as 3, when it describes a fountain in Solomon's temple. I think the proof of the irrationality of $pi$ had to wait for Lambert in the 1700s.



    Irrational numbers came about to solve problems that rational numbers were not up to the task to do. Early mathematicians resisted the concept, but they have been well accepted for the past millennium at least.






    share|cite|improve this answer



















    • 1




      Here is Archimedes' method of calculating the value of $pi$.
      – Yashbhatt
      Jun 30 '14 at 15:56












    • @David thanks for the edit! Can't believe I missed that.
      – Joel
      Jul 1 '14 at 13:49


















    up vote
    1
    down vote













    Some relevant definitions seem to be in order.



    Real number: Any number on the continuous real line from $-infty$ to $infty$.



    Integer: A real number which can be expressed without a fractional component.



    Rational number: A number which can be expressed as a ratio of two integers. Note that integers themselves are rational, since we can express any integer $n$ as $frac{n}{1}$.



    Irrational number: A number which cannot be expressed as a ratio of two integers.



    Now, in particular, in your question $kcdot10^{-n}$ is $textit{not}$ an integer where $k$ and $n$ are positive integers and $k$ is a digit between $1$ and $9$. These are instead just $textit{rational}$ numbers since $kcdot10^{-n}=dfrac{k}{10^n}$, which is a ratio of integers.



    Upsettingly for Pythagoras, irrational numbers do exist. Here is the standard proof that $sqrt{2}$ is irrational.






    share|cite|improve this answer




























      up vote
      0
      down vote













      Because you can not represent them as p/q with both p and q integers. That is why they are called irrational numbers. Not representable as p/q. 0.33333....is 1/3. root 2 can not be written as p/q where p and q are integers.






      share|cite|improve this answer






























        up vote
        0
        down vote













        Numbers that cannot be expressed rationally arise naturally as solutions of equations. The solution of $x^2=2$ & of $x(1-x)=1$, & any of an infinitude of specificable polynomial equations, cannot be expressed rationally. It can be proven directly, by a simple recipe, that no rational number can satisfy either of the two equations I have particularly cited; and the same can be proven similarly for the general case.



        Moreover there is a class of numbers yet beyond this: the transcendental numbers, any of which can be shown to be not the solution of any polynomial equation comprising a finite number of terms having integer coefficients.






        share|cite|improve this answer





















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          5 Answers
          5






          active

          oldest

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          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

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          active

          oldest

          votes








          up vote
          7
          down vote













          You can represent any real number as some convergent sequence of rational numbers, as you do above. However, irrational numbers are those numbers that cannot be expressed as any finite such sequence. However long of a finite sequence of rational numbers approaching $pi$ you have, there is a positive real number $varepsilon$ such that the last term in the sequence is at least $varepsilon$ away from $pi$ (In particular, we can take the last term of the sequence, $t$, and take $varepsilon = frac{|pi - t|}{2}$).






          share|cite|improve this answer

























            up vote
            7
            down vote













            You can represent any real number as some convergent sequence of rational numbers, as you do above. However, irrational numbers are those numbers that cannot be expressed as any finite such sequence. However long of a finite sequence of rational numbers approaching $pi$ you have, there is a positive real number $varepsilon$ such that the last term in the sequence is at least $varepsilon$ away from $pi$ (In particular, we can take the last term of the sequence, $t$, and take $varepsilon = frac{|pi - t|}{2}$).






            share|cite|improve this answer























              up vote
              7
              down vote










              up vote
              7
              down vote









              You can represent any real number as some convergent sequence of rational numbers, as you do above. However, irrational numbers are those numbers that cannot be expressed as any finite such sequence. However long of a finite sequence of rational numbers approaching $pi$ you have, there is a positive real number $varepsilon$ such that the last term in the sequence is at least $varepsilon$ away from $pi$ (In particular, we can take the last term of the sequence, $t$, and take $varepsilon = frac{|pi - t|}{2}$).






              share|cite|improve this answer












              You can represent any real number as some convergent sequence of rational numbers, as you do above. However, irrational numbers are those numbers that cannot be expressed as any finite such sequence. However long of a finite sequence of rational numbers approaching $pi$ you have, there is a positive real number $varepsilon$ such that the last term in the sequence is at least $varepsilon$ away from $pi$ (In particular, we can take the last term of the sequence, $t$, and take $varepsilon = frac{|pi - t|}{2}$).







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jun 24 '14 at 18:50









              qaphla

              3,492919




              3,492919






















                  up vote
                  5
                  down vote













                  For a long time people believed that the rational numbers were enough to perform any arithmetic operation we needed. It wasn't until 500 BC that the Pythagoreans started to become aware that the rational numbers were not quite sufficient.



                  For instance, the Pythagoreans tried to compute something that seemed benign. They wanted to find the length of the diagonal of a square with sides of unit length. We know now (by the Pythagorean theorem) that this should be the number we represent as $sqrt{2}$. The Pythagoreans tried to find a rational number corresponding to this and came to a contradiction. It's said that the Pythagoreans were so upset that they drowned the person who discovered this. I have also heard speculation that this is why Greek mathematics was so focused on geometry, since this is something you can "see" and away from uncomfortable concepts like irrational numbers.



                  This was the first instance where irrational number started to seem necessary. The next big irrational number to come forward was the number $pi$. This was in 350 BC, and the first to try to approximate this number was Archimedes. This is where the estimate 22/7 came from. He also had a better estimate than that, it escapes me at the moment. The Bible also estimates this as 3, when it describes a fountain in Solomon's temple. I think the proof of the irrationality of $pi$ had to wait for Lambert in the 1700s.



                  Irrational numbers came about to solve problems that rational numbers were not up to the task to do. Early mathematicians resisted the concept, but they have been well accepted for the past millennium at least.






                  share|cite|improve this answer



















                  • 1




                    Here is Archimedes' method of calculating the value of $pi$.
                    – Yashbhatt
                    Jun 30 '14 at 15:56












                  • @David thanks for the edit! Can't believe I missed that.
                    – Joel
                    Jul 1 '14 at 13:49















                  up vote
                  5
                  down vote













                  For a long time people believed that the rational numbers were enough to perform any arithmetic operation we needed. It wasn't until 500 BC that the Pythagoreans started to become aware that the rational numbers were not quite sufficient.



                  For instance, the Pythagoreans tried to compute something that seemed benign. They wanted to find the length of the diagonal of a square with sides of unit length. We know now (by the Pythagorean theorem) that this should be the number we represent as $sqrt{2}$. The Pythagoreans tried to find a rational number corresponding to this and came to a contradiction. It's said that the Pythagoreans were so upset that they drowned the person who discovered this. I have also heard speculation that this is why Greek mathematics was so focused on geometry, since this is something you can "see" and away from uncomfortable concepts like irrational numbers.



                  This was the first instance where irrational number started to seem necessary. The next big irrational number to come forward was the number $pi$. This was in 350 BC, and the first to try to approximate this number was Archimedes. This is where the estimate 22/7 came from. He also had a better estimate than that, it escapes me at the moment. The Bible also estimates this as 3, when it describes a fountain in Solomon's temple. I think the proof of the irrationality of $pi$ had to wait for Lambert in the 1700s.



                  Irrational numbers came about to solve problems that rational numbers were not up to the task to do. Early mathematicians resisted the concept, but they have been well accepted for the past millennium at least.






                  share|cite|improve this answer



















                  • 1




                    Here is Archimedes' method of calculating the value of $pi$.
                    – Yashbhatt
                    Jun 30 '14 at 15:56












                  • @David thanks for the edit! Can't believe I missed that.
                    – Joel
                    Jul 1 '14 at 13:49













                  up vote
                  5
                  down vote










                  up vote
                  5
                  down vote









                  For a long time people believed that the rational numbers were enough to perform any arithmetic operation we needed. It wasn't until 500 BC that the Pythagoreans started to become aware that the rational numbers were not quite sufficient.



                  For instance, the Pythagoreans tried to compute something that seemed benign. They wanted to find the length of the diagonal of a square with sides of unit length. We know now (by the Pythagorean theorem) that this should be the number we represent as $sqrt{2}$. The Pythagoreans tried to find a rational number corresponding to this and came to a contradiction. It's said that the Pythagoreans were so upset that they drowned the person who discovered this. I have also heard speculation that this is why Greek mathematics was so focused on geometry, since this is something you can "see" and away from uncomfortable concepts like irrational numbers.



                  This was the first instance where irrational number started to seem necessary. The next big irrational number to come forward was the number $pi$. This was in 350 BC, and the first to try to approximate this number was Archimedes. This is where the estimate 22/7 came from. He also had a better estimate than that, it escapes me at the moment. The Bible also estimates this as 3, when it describes a fountain in Solomon's temple. I think the proof of the irrationality of $pi$ had to wait for Lambert in the 1700s.



                  Irrational numbers came about to solve problems that rational numbers were not up to the task to do. Early mathematicians resisted the concept, but they have been well accepted for the past millennium at least.






                  share|cite|improve this answer














                  For a long time people believed that the rational numbers were enough to perform any arithmetic operation we needed. It wasn't until 500 BC that the Pythagoreans started to become aware that the rational numbers were not quite sufficient.



                  For instance, the Pythagoreans tried to compute something that seemed benign. They wanted to find the length of the diagonal of a square with sides of unit length. We know now (by the Pythagorean theorem) that this should be the number we represent as $sqrt{2}$. The Pythagoreans tried to find a rational number corresponding to this and came to a contradiction. It's said that the Pythagoreans were so upset that they drowned the person who discovered this. I have also heard speculation that this is why Greek mathematics was so focused on geometry, since this is something you can "see" and away from uncomfortable concepts like irrational numbers.



                  This was the first instance where irrational number started to seem necessary. The next big irrational number to come forward was the number $pi$. This was in 350 BC, and the first to try to approximate this number was Archimedes. This is where the estimate 22/7 came from. He also had a better estimate than that, it escapes me at the moment. The Bible also estimates this as 3, when it describes a fountain in Solomon's temple. I think the proof of the irrationality of $pi$ had to wait for Lambert in the 1700s.



                  Irrational numbers came about to solve problems that rational numbers were not up to the task to do. Early mathematicians resisted the concept, but they have been well accepted for the past millennium at least.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jul 1 '14 at 5:20









                  David

                  67.4k663126




                  67.4k663126










                  answered Jun 25 '14 at 14:40









                  Joel

                  14.2k12240




                  14.2k12240








                  • 1




                    Here is Archimedes' method of calculating the value of $pi$.
                    – Yashbhatt
                    Jun 30 '14 at 15:56












                  • @David thanks for the edit! Can't believe I missed that.
                    – Joel
                    Jul 1 '14 at 13:49














                  • 1




                    Here is Archimedes' method of calculating the value of $pi$.
                    – Yashbhatt
                    Jun 30 '14 at 15:56












                  • @David thanks for the edit! Can't believe I missed that.
                    – Joel
                    Jul 1 '14 at 13:49








                  1




                  1




                  Here is Archimedes' method of calculating the value of $pi$.
                  – Yashbhatt
                  Jun 30 '14 at 15:56






                  Here is Archimedes' method of calculating the value of $pi$.
                  – Yashbhatt
                  Jun 30 '14 at 15:56














                  @David thanks for the edit! Can't believe I missed that.
                  – Joel
                  Jul 1 '14 at 13:49




                  @David thanks for the edit! Can't believe I missed that.
                  – Joel
                  Jul 1 '14 at 13:49










                  up vote
                  1
                  down vote













                  Some relevant definitions seem to be in order.



                  Real number: Any number on the continuous real line from $-infty$ to $infty$.



                  Integer: A real number which can be expressed without a fractional component.



                  Rational number: A number which can be expressed as a ratio of two integers. Note that integers themselves are rational, since we can express any integer $n$ as $frac{n}{1}$.



                  Irrational number: A number which cannot be expressed as a ratio of two integers.



                  Now, in particular, in your question $kcdot10^{-n}$ is $textit{not}$ an integer where $k$ and $n$ are positive integers and $k$ is a digit between $1$ and $9$. These are instead just $textit{rational}$ numbers since $kcdot10^{-n}=dfrac{k}{10^n}$, which is a ratio of integers.



                  Upsettingly for Pythagoras, irrational numbers do exist. Here is the standard proof that $sqrt{2}$ is irrational.






                  share|cite|improve this answer

























                    up vote
                    1
                    down vote













                    Some relevant definitions seem to be in order.



                    Real number: Any number on the continuous real line from $-infty$ to $infty$.



                    Integer: A real number which can be expressed without a fractional component.



                    Rational number: A number which can be expressed as a ratio of two integers. Note that integers themselves are rational, since we can express any integer $n$ as $frac{n}{1}$.



                    Irrational number: A number which cannot be expressed as a ratio of two integers.



                    Now, in particular, in your question $kcdot10^{-n}$ is $textit{not}$ an integer where $k$ and $n$ are positive integers and $k$ is a digit between $1$ and $9$. These are instead just $textit{rational}$ numbers since $kcdot10^{-n}=dfrac{k}{10^n}$, which is a ratio of integers.



                    Upsettingly for Pythagoras, irrational numbers do exist. Here is the standard proof that $sqrt{2}$ is irrational.






                    share|cite|improve this answer























                      up vote
                      1
                      down vote










                      up vote
                      1
                      down vote









                      Some relevant definitions seem to be in order.



                      Real number: Any number on the continuous real line from $-infty$ to $infty$.



                      Integer: A real number which can be expressed without a fractional component.



                      Rational number: A number which can be expressed as a ratio of two integers. Note that integers themselves are rational, since we can express any integer $n$ as $frac{n}{1}$.



                      Irrational number: A number which cannot be expressed as a ratio of two integers.



                      Now, in particular, in your question $kcdot10^{-n}$ is $textit{not}$ an integer where $k$ and $n$ are positive integers and $k$ is a digit between $1$ and $9$. These are instead just $textit{rational}$ numbers since $kcdot10^{-n}=dfrac{k}{10^n}$, which is a ratio of integers.



                      Upsettingly for Pythagoras, irrational numbers do exist. Here is the standard proof that $sqrt{2}$ is irrational.






                      share|cite|improve this answer












                      Some relevant definitions seem to be in order.



                      Real number: Any number on the continuous real line from $-infty$ to $infty$.



                      Integer: A real number which can be expressed without a fractional component.



                      Rational number: A number which can be expressed as a ratio of two integers. Note that integers themselves are rational, since we can express any integer $n$ as $frac{n}{1}$.



                      Irrational number: A number which cannot be expressed as a ratio of two integers.



                      Now, in particular, in your question $kcdot10^{-n}$ is $textit{not}$ an integer where $k$ and $n$ are positive integers and $k$ is a digit between $1$ and $9$. These are instead just $textit{rational}$ numbers since $kcdot10^{-n}=dfrac{k}{10^n}$, which is a ratio of integers.



                      Upsettingly for Pythagoras, irrational numbers do exist. Here is the standard proof that $sqrt{2}$ is irrational.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jun 24 '14 at 19:15









                      Peter Woolfitt

                      18.5k54478




                      18.5k54478






















                          up vote
                          0
                          down vote













                          Because you can not represent them as p/q with both p and q integers. That is why they are called irrational numbers. Not representable as p/q. 0.33333....is 1/3. root 2 can not be written as p/q where p and q are integers.






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                            up vote
                            0
                            down vote













                            Because you can not represent them as p/q with both p and q integers. That is why they are called irrational numbers. Not representable as p/q. 0.33333....is 1/3. root 2 can not be written as p/q where p and q are integers.






                            share|cite|improve this answer

























                              up vote
                              0
                              down vote










                              up vote
                              0
                              down vote









                              Because you can not represent them as p/q with both p and q integers. That is why they are called irrational numbers. Not representable as p/q. 0.33333....is 1/3. root 2 can not be written as p/q where p and q are integers.






                              share|cite|improve this answer














                              Because you can not represent them as p/q with both p and q integers. That is why they are called irrational numbers. Not representable as p/q. 0.33333....is 1/3. root 2 can not be written as p/q where p and q are integers.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Jun 30 '14 at 15:41









                              Yashbhatt

                              1238




                              1238










                              answered Jun 24 '14 at 18:52









                              Seetha Rama Raju Sanapala

                              8319




                              8319






















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                                  Numbers that cannot be expressed rationally arise naturally as solutions of equations. The solution of $x^2=2$ & of $x(1-x)=1$, & any of an infinitude of specificable polynomial equations, cannot be expressed rationally. It can be proven directly, by a simple recipe, that no rational number can satisfy either of the two equations I have particularly cited; and the same can be proven similarly for the general case.



                                  Moreover there is a class of numbers yet beyond this: the transcendental numbers, any of which can be shown to be not the solution of any polynomial equation comprising a finite number of terms having integer coefficients.






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                                    down vote













                                    Numbers that cannot be expressed rationally arise naturally as solutions of equations. The solution of $x^2=2$ & of $x(1-x)=1$, & any of an infinitude of specificable polynomial equations, cannot be expressed rationally. It can be proven directly, by a simple recipe, that no rational number can satisfy either of the two equations I have particularly cited; and the same can be proven similarly for the general case.



                                    Moreover there is a class of numbers yet beyond this: the transcendental numbers, any of which can be shown to be not the solution of any polynomial equation comprising a finite number of terms having integer coefficients.






                                    share|cite|improve this answer























                                      up vote
                                      0
                                      down vote










                                      up vote
                                      0
                                      down vote









                                      Numbers that cannot be expressed rationally arise naturally as solutions of equations. The solution of $x^2=2$ & of $x(1-x)=1$, & any of an infinitude of specificable polynomial equations, cannot be expressed rationally. It can be proven directly, by a simple recipe, that no rational number can satisfy either of the two equations I have particularly cited; and the same can be proven similarly for the general case.



                                      Moreover there is a class of numbers yet beyond this: the transcendental numbers, any of which can be shown to be not the solution of any polynomial equation comprising a finite number of terms having integer coefficients.






                                      share|cite|improve this answer












                                      Numbers that cannot be expressed rationally arise naturally as solutions of equations. The solution of $x^2=2$ & of $x(1-x)=1$, & any of an infinitude of specificable polynomial equations, cannot be expressed rationally. It can be proven directly, by a simple recipe, that no rational number can satisfy either of the two equations I have particularly cited; and the same can be proven similarly for the general case.



                                      Moreover there is a class of numbers yet beyond this: the transcendental numbers, any of which can be shown to be not the solution of any polynomial equation comprising a finite number of terms having integer coefficients.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 22 at 5:05









                                      AmbretteOrrisey

                                      49110




                                      49110






























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