Is a number written in the square root/fraction form called a non-integer even if it can be simplified to an...












3














This is very simple question, but I cannot get the ansewer from the internet.



Is a number written in the square root/fraction form called a non-integer even if it can be simplified to an integer.



For example 4/2, 12/4, sqr4, sqr64 etc... do these need to be simplified before we can call them integers.



Too make this easer to understand are sqr64 and 12/4 non-integers while 8 and 3 are integers.










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closed as unclear what you're asking by Lord Shark the Unknown, KReiser, Cesareo, Leucippus, Rebellos Nov 28 '18 at 9:27


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.











  • 1




    $frac 42=2$ so that's an integer, however you have written it. Similarly for the others.
    – lulu
    Nov 27 '18 at 21:05












  • Integers is a subset of Rational numbers, Of course an integer is also a rational number. So the point is no!
    – mathnoob
    Nov 27 '18 at 21:09










  • Thank you to all.
    – Asan Ramzan
    Nov 27 '18 at 21:26










  • I know this put on hold because its unclear, but everyone who has answered this question seems to understand what I am saying, so I don't see where the problem is. Could you elaborate
    – Asan Ramzan
    Nov 28 '18 at 12:03










  • @AsanRamzan I'm surprised the question was put on hold; I've voted to reopen it. Do you feel you have a better understanding after reading the answers?
    – Théophile
    Nov 28 '18 at 16:45
















3














This is very simple question, but I cannot get the ansewer from the internet.



Is a number written in the square root/fraction form called a non-integer even if it can be simplified to an integer.



For example 4/2, 12/4, sqr4, sqr64 etc... do these need to be simplified before we can call them integers.



Too make this easer to understand are sqr64 and 12/4 non-integers while 8 and 3 are integers.










share|cite|improve this question















closed as unclear what you're asking by Lord Shark the Unknown, KReiser, Cesareo, Leucippus, Rebellos Nov 28 '18 at 9:27


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.











  • 1




    $frac 42=2$ so that's an integer, however you have written it. Similarly for the others.
    – lulu
    Nov 27 '18 at 21:05












  • Integers is a subset of Rational numbers, Of course an integer is also a rational number. So the point is no!
    – mathnoob
    Nov 27 '18 at 21:09










  • Thank you to all.
    – Asan Ramzan
    Nov 27 '18 at 21:26










  • I know this put on hold because its unclear, but everyone who has answered this question seems to understand what I am saying, so I don't see where the problem is. Could you elaborate
    – Asan Ramzan
    Nov 28 '18 at 12:03










  • @AsanRamzan I'm surprised the question was put on hold; I've voted to reopen it. Do you feel you have a better understanding after reading the answers?
    – Théophile
    Nov 28 '18 at 16:45














3












3








3







This is very simple question, but I cannot get the ansewer from the internet.



Is a number written in the square root/fraction form called a non-integer even if it can be simplified to an integer.



For example 4/2, 12/4, sqr4, sqr64 etc... do these need to be simplified before we can call them integers.



Too make this easer to understand are sqr64 and 12/4 non-integers while 8 and 3 are integers.










share|cite|improve this question















This is very simple question, but I cannot get the ansewer from the internet.



Is a number written in the square root/fraction form called a non-integer even if it can be simplified to an integer.



For example 4/2, 12/4, sqr4, sqr64 etc... do these need to be simplified before we can call them integers.



Too make this easer to understand are sqr64 and 12/4 non-integers while 8 and 3 are integers.







square-numbers






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share|cite|improve this question













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share|cite|improve this question








edited Nov 28 '18 at 16:49

























asked Nov 27 '18 at 21:03









Asan Ramzan

162




162




closed as unclear what you're asking by Lord Shark the Unknown, KReiser, Cesareo, Leucippus, Rebellos Nov 28 '18 at 9:27


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






closed as unclear what you're asking by Lord Shark the Unknown, KReiser, Cesareo, Leucippus, Rebellos Nov 28 '18 at 9:27


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.










  • 1




    $frac 42=2$ so that's an integer, however you have written it. Similarly for the others.
    – lulu
    Nov 27 '18 at 21:05












  • Integers is a subset of Rational numbers, Of course an integer is also a rational number. So the point is no!
    – mathnoob
    Nov 27 '18 at 21:09










  • Thank you to all.
    – Asan Ramzan
    Nov 27 '18 at 21:26










  • I know this put on hold because its unclear, but everyone who has answered this question seems to understand what I am saying, so I don't see where the problem is. Could you elaborate
    – Asan Ramzan
    Nov 28 '18 at 12:03










  • @AsanRamzan I'm surprised the question was put on hold; I've voted to reopen it. Do you feel you have a better understanding after reading the answers?
    – Théophile
    Nov 28 '18 at 16:45














  • 1




    $frac 42=2$ so that's an integer, however you have written it. Similarly for the others.
    – lulu
    Nov 27 '18 at 21:05












  • Integers is a subset of Rational numbers, Of course an integer is also a rational number. So the point is no!
    – mathnoob
    Nov 27 '18 at 21:09










  • Thank you to all.
    – Asan Ramzan
    Nov 27 '18 at 21:26










  • I know this put on hold because its unclear, but everyone who has answered this question seems to understand what I am saying, so I don't see where the problem is. Could you elaborate
    – Asan Ramzan
    Nov 28 '18 at 12:03










  • @AsanRamzan I'm surprised the question was put on hold; I've voted to reopen it. Do you feel you have a better understanding after reading the answers?
    – Théophile
    Nov 28 '18 at 16:45








1




1




$frac 42=2$ so that's an integer, however you have written it. Similarly for the others.
– lulu
Nov 27 '18 at 21:05






$frac 42=2$ so that's an integer, however you have written it. Similarly for the others.
– lulu
Nov 27 '18 at 21:05














Integers is a subset of Rational numbers, Of course an integer is also a rational number. So the point is no!
– mathnoob
Nov 27 '18 at 21:09




Integers is a subset of Rational numbers, Of course an integer is also a rational number. So the point is no!
– mathnoob
Nov 27 '18 at 21:09












Thank you to all.
– Asan Ramzan
Nov 27 '18 at 21:26




Thank you to all.
– Asan Ramzan
Nov 27 '18 at 21:26












I know this put on hold because its unclear, but everyone who has answered this question seems to understand what I am saying, so I don't see where the problem is. Could you elaborate
– Asan Ramzan
Nov 28 '18 at 12:03




I know this put on hold because its unclear, but everyone who has answered this question seems to understand what I am saying, so I don't see where the problem is. Could you elaborate
– Asan Ramzan
Nov 28 '18 at 12:03












@AsanRamzan I'm surprised the question was put on hold; I've voted to reopen it. Do you feel you have a better understanding after reading the answers?
– Théophile
Nov 28 '18 at 16:45




@AsanRamzan I'm surprised the question was put on hold; I've voted to reopen it. Do you feel you have a better understanding after reading the answers?
– Théophile
Nov 28 '18 at 16:45










4 Answers
4






active

oldest

votes


















4














No. Numbers are what they are. It doesn't matter how they are represented.



$7$ is an integer. Period.



It doesn't matter if is written as $5 + 2$ or $sqrt{49}$ or $sqrt{25} + frac{sqrt[3]{16}}{2^{frac 13}}$ or $ln (e^7)$.



Those are all equal to $7$ and $7$ is an integer. Period.



====



That said, it might not be easy (or even possible) to tell if a number is or is not an integer. It's obvious that $7$ is an integer and $7.0000012142650469991421281354411.....$ isn't. But it isn't clear whether $sqrt[7]{823543}$ or $sqrt[7]{823544}$ are integers. (It turns out that those are the same numbers.)



But it doesn't matter whether we know if a number is an integer or not. It either is or isn't.






share|cite|improve this answer































    1














    It is helpful to have vocabulary to distinguish between what an object fundamentally is and how it is represented.



    A fraction $a/b$ is said to be irreducible or in simplest form if $a$ and $b$ are integers with no common factors, i.e., $gcd(a,b) = 1$. So, for example, $2/3$ is irreducible, while $200/300$ is not.



    An integer can be written down in many ways. Not all of those ways are simple, but fundamentally the number is the same. Take $2$, for instance. We could write it as $10/5$: this is not the simplest form of the fraction, but the number is still an integer. Or we could express it as
    $$1 + 1/2 + 1/4 + 1/8 + cdots;$$
    again, this is not as straightforward as simply writing "$2$", but the expression still represents an integer.






    share|cite|improve this answer





























      1














      $2$ is an integer. $4/2$ is equal to $2$, and therefore has all the properties that the number $2$ has, including being an integer. The square root of $4$ is also equal to $2$, so it's an integer as well. In some cases, you'll probably need to simplify to recognize that it is indeed an integer, but that doesn't change its properties no matter how you write it.



      For example, is $sqrt{14883}$ an integer? How about $sqrt{14884}$? It might be tough to tell unless you do the simplification, but one is an integer and one isn't.






      share|cite|improve this answer































        -4














        If it can be simplified to an integer,
        it can be called an integer
        after the simplification.



        Until the simplification is done,
        I would just call the expression
        "an expression"
        when it is not clear if it could be
        simplified to an integer.



        Considering how expressions
        involving nested radicals
        can be sometimes
        amazingly simplified,
        I think that there would be cases
        where the fact that
        an expression simplifies
        to an integer
        is a surprise.






        share|cite|improve this answer





















        • I disagree. The number either is or is not an integer whether we know whether it is or is not. So it IS an integer before we simplify it. It's just that before we simplify it we don't KNOW if it is an integer. But that doesn't mean it isn't an integer. (because ... it is.)
          – fleablood
          Nov 27 '18 at 21:22










        • Suppose the expression encodes something we don't know the answer for - for example, 1 if $pi$ is a normal number, 1/2 if it is not. Is the value of the expression an integer?
          – marty cohen
          Nov 27 '18 at 23:38






        • 1




          " Is the value of the expression an integer?" If $pi$ is normal then it is an integer. If $pi$ is not normal it is not an integer. The fact that we don't know if it's an integer or not doesn't prevent it from being an integer or a non-integer.
          – fleablood
          Nov 28 '18 at 0:04










        • No. A normal number is a real number: en.wikipedia.org/wiki/Normal_number
          – marty cohen
          Nov 28 '18 at 3:47






        • 1




          Um.... What are you talking about??? Of course normal numbers are real numbers. Everything that isnt complex is a real number. You asked if given the function $f(x) = 1$ if $x$ is normal; $f(x)$ if $x$ is not normal. You asked whether $f(pi)$ is an integer (which it is if it is $1$) or if it was not an integer (which it will not be if it is $frac 12$). And the answer is: We don't know. But EITHER $f(pi)$ is an integer OR it is not. We can NOT say $f(pi)$ is a non-integer, just because we don't know that it is an integer. We don't know it is not.
          – fleablood
          Nov 28 '18 at 5:38


















        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4














        No. Numbers are what they are. It doesn't matter how they are represented.



        $7$ is an integer. Period.



        It doesn't matter if is written as $5 + 2$ or $sqrt{49}$ or $sqrt{25} + frac{sqrt[3]{16}}{2^{frac 13}}$ or $ln (e^7)$.



        Those are all equal to $7$ and $7$ is an integer. Period.



        ====



        That said, it might not be easy (or even possible) to tell if a number is or is not an integer. It's obvious that $7$ is an integer and $7.0000012142650469991421281354411.....$ isn't. But it isn't clear whether $sqrt[7]{823543}$ or $sqrt[7]{823544}$ are integers. (It turns out that those are the same numbers.)



        But it doesn't matter whether we know if a number is an integer or not. It either is or isn't.






        share|cite|improve this answer




























          4














          No. Numbers are what they are. It doesn't matter how they are represented.



          $7$ is an integer. Period.



          It doesn't matter if is written as $5 + 2$ or $sqrt{49}$ or $sqrt{25} + frac{sqrt[3]{16}}{2^{frac 13}}$ or $ln (e^7)$.



          Those are all equal to $7$ and $7$ is an integer. Period.



          ====



          That said, it might not be easy (or even possible) to tell if a number is or is not an integer. It's obvious that $7$ is an integer and $7.0000012142650469991421281354411.....$ isn't. But it isn't clear whether $sqrt[7]{823543}$ or $sqrt[7]{823544}$ are integers. (It turns out that those are the same numbers.)



          But it doesn't matter whether we know if a number is an integer or not. It either is or isn't.






          share|cite|improve this answer


























            4












            4








            4






            No. Numbers are what they are. It doesn't matter how they are represented.



            $7$ is an integer. Period.



            It doesn't matter if is written as $5 + 2$ or $sqrt{49}$ or $sqrt{25} + frac{sqrt[3]{16}}{2^{frac 13}}$ or $ln (e^7)$.



            Those are all equal to $7$ and $7$ is an integer. Period.



            ====



            That said, it might not be easy (or even possible) to tell if a number is or is not an integer. It's obvious that $7$ is an integer and $7.0000012142650469991421281354411.....$ isn't. But it isn't clear whether $sqrt[7]{823543}$ or $sqrt[7]{823544}$ are integers. (It turns out that those are the same numbers.)



            But it doesn't matter whether we know if a number is an integer or not. It either is or isn't.






            share|cite|improve this answer














            No. Numbers are what they are. It doesn't matter how they are represented.



            $7$ is an integer. Period.



            It doesn't matter if is written as $5 + 2$ or $sqrt{49}$ or $sqrt{25} + frac{sqrt[3]{16}}{2^{frac 13}}$ or $ln (e^7)$.



            Those are all equal to $7$ and $7$ is an integer. Period.



            ====



            That said, it might not be easy (or even possible) to tell if a number is or is not an integer. It's obvious that $7$ is an integer and $7.0000012142650469991421281354411.....$ isn't. But it isn't clear whether $sqrt[7]{823543}$ or $sqrt[7]{823544}$ are integers. (It turns out that those are the same numbers.)



            But it doesn't matter whether we know if a number is an integer or not. It either is or isn't.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 27 '18 at 21:27

























            answered Nov 27 '18 at 21:20









            fleablood

            68.3k22685




            68.3k22685























                1














                It is helpful to have vocabulary to distinguish between what an object fundamentally is and how it is represented.



                A fraction $a/b$ is said to be irreducible or in simplest form if $a$ and $b$ are integers with no common factors, i.e., $gcd(a,b) = 1$. So, for example, $2/3$ is irreducible, while $200/300$ is not.



                An integer can be written down in many ways. Not all of those ways are simple, but fundamentally the number is the same. Take $2$, for instance. We could write it as $10/5$: this is not the simplest form of the fraction, but the number is still an integer. Or we could express it as
                $$1 + 1/2 + 1/4 + 1/8 + cdots;$$
                again, this is not as straightforward as simply writing "$2$", but the expression still represents an integer.






                share|cite|improve this answer


























                  1














                  It is helpful to have vocabulary to distinguish between what an object fundamentally is and how it is represented.



                  A fraction $a/b$ is said to be irreducible or in simplest form if $a$ and $b$ are integers with no common factors, i.e., $gcd(a,b) = 1$. So, for example, $2/3$ is irreducible, while $200/300$ is not.



                  An integer can be written down in many ways. Not all of those ways are simple, but fundamentally the number is the same. Take $2$, for instance. We could write it as $10/5$: this is not the simplest form of the fraction, but the number is still an integer. Or we could express it as
                  $$1 + 1/2 + 1/4 + 1/8 + cdots;$$
                  again, this is not as straightforward as simply writing "$2$", but the expression still represents an integer.






                  share|cite|improve this answer
























                    1












                    1








                    1






                    It is helpful to have vocabulary to distinguish between what an object fundamentally is and how it is represented.



                    A fraction $a/b$ is said to be irreducible or in simplest form if $a$ and $b$ are integers with no common factors, i.e., $gcd(a,b) = 1$. So, for example, $2/3$ is irreducible, while $200/300$ is not.



                    An integer can be written down in many ways. Not all of those ways are simple, but fundamentally the number is the same. Take $2$, for instance. We could write it as $10/5$: this is not the simplest form of the fraction, but the number is still an integer. Or we could express it as
                    $$1 + 1/2 + 1/4 + 1/8 + cdots;$$
                    again, this is not as straightforward as simply writing "$2$", but the expression still represents an integer.






                    share|cite|improve this answer












                    It is helpful to have vocabulary to distinguish between what an object fundamentally is and how it is represented.



                    A fraction $a/b$ is said to be irreducible or in simplest form if $a$ and $b$ are integers with no common factors, i.e., $gcd(a,b) = 1$. So, for example, $2/3$ is irreducible, while $200/300$ is not.



                    An integer can be written down in many ways. Not all of those ways are simple, but fundamentally the number is the same. Take $2$, for instance. We could write it as $10/5$: this is not the simplest form of the fraction, but the number is still an integer. Or we could express it as
                    $$1 + 1/2 + 1/4 + 1/8 + cdots;$$
                    again, this is not as straightforward as simply writing "$2$", but the expression still represents an integer.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 27 '18 at 21:16









                    Théophile

                    19.5k12946




                    19.5k12946























                        1














                        $2$ is an integer. $4/2$ is equal to $2$, and therefore has all the properties that the number $2$ has, including being an integer. The square root of $4$ is also equal to $2$, so it's an integer as well. In some cases, you'll probably need to simplify to recognize that it is indeed an integer, but that doesn't change its properties no matter how you write it.



                        For example, is $sqrt{14883}$ an integer? How about $sqrt{14884}$? It might be tough to tell unless you do the simplification, but one is an integer and one isn't.






                        share|cite|improve this answer




























                          1














                          $2$ is an integer. $4/2$ is equal to $2$, and therefore has all the properties that the number $2$ has, including being an integer. The square root of $4$ is also equal to $2$, so it's an integer as well. In some cases, you'll probably need to simplify to recognize that it is indeed an integer, but that doesn't change its properties no matter how you write it.



                          For example, is $sqrt{14883}$ an integer? How about $sqrt{14884}$? It might be tough to tell unless you do the simplification, but one is an integer and one isn't.






                          share|cite|improve this answer


























                            1












                            1








                            1






                            $2$ is an integer. $4/2$ is equal to $2$, and therefore has all the properties that the number $2$ has, including being an integer. The square root of $4$ is also equal to $2$, so it's an integer as well. In some cases, you'll probably need to simplify to recognize that it is indeed an integer, but that doesn't change its properties no matter how you write it.



                            For example, is $sqrt{14883}$ an integer? How about $sqrt{14884}$? It might be tough to tell unless you do the simplification, but one is an integer and one isn't.






                            share|cite|improve this answer














                            $2$ is an integer. $4/2$ is equal to $2$, and therefore has all the properties that the number $2$ has, including being an integer. The square root of $4$ is also equal to $2$, so it's an integer as well. In some cases, you'll probably need to simplify to recognize that it is indeed an integer, but that doesn't change its properties no matter how you write it.



                            For example, is $sqrt{14883}$ an integer? How about $sqrt{14884}$? It might be tough to tell unless you do the simplification, but one is an integer and one isn't.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Nov 27 '18 at 21:17









                            Théophile

                            19.5k12946




                            19.5k12946










                            answered Nov 27 '18 at 21:12









                            Nuclear Wang

                            41127




                            41127























                                -4














                                If it can be simplified to an integer,
                                it can be called an integer
                                after the simplification.



                                Until the simplification is done,
                                I would just call the expression
                                "an expression"
                                when it is not clear if it could be
                                simplified to an integer.



                                Considering how expressions
                                involving nested radicals
                                can be sometimes
                                amazingly simplified,
                                I think that there would be cases
                                where the fact that
                                an expression simplifies
                                to an integer
                                is a surprise.






                                share|cite|improve this answer





















                                • I disagree. The number either is or is not an integer whether we know whether it is or is not. So it IS an integer before we simplify it. It's just that before we simplify it we don't KNOW if it is an integer. But that doesn't mean it isn't an integer. (because ... it is.)
                                  – fleablood
                                  Nov 27 '18 at 21:22










                                • Suppose the expression encodes something we don't know the answer for - for example, 1 if $pi$ is a normal number, 1/2 if it is not. Is the value of the expression an integer?
                                  – marty cohen
                                  Nov 27 '18 at 23:38






                                • 1




                                  " Is the value of the expression an integer?" If $pi$ is normal then it is an integer. If $pi$ is not normal it is not an integer. The fact that we don't know if it's an integer or not doesn't prevent it from being an integer or a non-integer.
                                  – fleablood
                                  Nov 28 '18 at 0:04










                                • No. A normal number is a real number: en.wikipedia.org/wiki/Normal_number
                                  – marty cohen
                                  Nov 28 '18 at 3:47






                                • 1




                                  Um.... What are you talking about??? Of course normal numbers are real numbers. Everything that isnt complex is a real number. You asked if given the function $f(x) = 1$ if $x$ is normal; $f(x)$ if $x$ is not normal. You asked whether $f(pi)$ is an integer (which it is if it is $1$) or if it was not an integer (which it will not be if it is $frac 12$). And the answer is: We don't know. But EITHER $f(pi)$ is an integer OR it is not. We can NOT say $f(pi)$ is a non-integer, just because we don't know that it is an integer. We don't know it is not.
                                  – fleablood
                                  Nov 28 '18 at 5:38
















                                -4














                                If it can be simplified to an integer,
                                it can be called an integer
                                after the simplification.



                                Until the simplification is done,
                                I would just call the expression
                                "an expression"
                                when it is not clear if it could be
                                simplified to an integer.



                                Considering how expressions
                                involving nested radicals
                                can be sometimes
                                amazingly simplified,
                                I think that there would be cases
                                where the fact that
                                an expression simplifies
                                to an integer
                                is a surprise.






                                share|cite|improve this answer





















                                • I disagree. The number either is or is not an integer whether we know whether it is or is not. So it IS an integer before we simplify it. It's just that before we simplify it we don't KNOW if it is an integer. But that doesn't mean it isn't an integer. (because ... it is.)
                                  – fleablood
                                  Nov 27 '18 at 21:22










                                • Suppose the expression encodes something we don't know the answer for - for example, 1 if $pi$ is a normal number, 1/2 if it is not. Is the value of the expression an integer?
                                  – marty cohen
                                  Nov 27 '18 at 23:38






                                • 1




                                  " Is the value of the expression an integer?" If $pi$ is normal then it is an integer. If $pi$ is not normal it is not an integer. The fact that we don't know if it's an integer or not doesn't prevent it from being an integer or a non-integer.
                                  – fleablood
                                  Nov 28 '18 at 0:04










                                • No. A normal number is a real number: en.wikipedia.org/wiki/Normal_number
                                  – marty cohen
                                  Nov 28 '18 at 3:47






                                • 1




                                  Um.... What are you talking about??? Of course normal numbers are real numbers. Everything that isnt complex is a real number. You asked if given the function $f(x) = 1$ if $x$ is normal; $f(x)$ if $x$ is not normal. You asked whether $f(pi)$ is an integer (which it is if it is $1$) or if it was not an integer (which it will not be if it is $frac 12$). And the answer is: We don't know. But EITHER $f(pi)$ is an integer OR it is not. We can NOT say $f(pi)$ is a non-integer, just because we don't know that it is an integer. We don't know it is not.
                                  – fleablood
                                  Nov 28 '18 at 5:38














                                -4












                                -4








                                -4






                                If it can be simplified to an integer,
                                it can be called an integer
                                after the simplification.



                                Until the simplification is done,
                                I would just call the expression
                                "an expression"
                                when it is not clear if it could be
                                simplified to an integer.



                                Considering how expressions
                                involving nested radicals
                                can be sometimes
                                amazingly simplified,
                                I think that there would be cases
                                where the fact that
                                an expression simplifies
                                to an integer
                                is a surprise.






                                share|cite|improve this answer












                                If it can be simplified to an integer,
                                it can be called an integer
                                after the simplification.



                                Until the simplification is done,
                                I would just call the expression
                                "an expression"
                                when it is not clear if it could be
                                simplified to an integer.



                                Considering how expressions
                                involving nested radicals
                                can be sometimes
                                amazingly simplified,
                                I think that there would be cases
                                where the fact that
                                an expression simplifies
                                to an integer
                                is a surprise.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Nov 27 '18 at 21:10









                                marty cohen

                                72.6k549127




                                72.6k549127












                                • I disagree. The number either is or is not an integer whether we know whether it is or is not. So it IS an integer before we simplify it. It's just that before we simplify it we don't KNOW if it is an integer. But that doesn't mean it isn't an integer. (because ... it is.)
                                  – fleablood
                                  Nov 27 '18 at 21:22










                                • Suppose the expression encodes something we don't know the answer for - for example, 1 if $pi$ is a normal number, 1/2 if it is not. Is the value of the expression an integer?
                                  – marty cohen
                                  Nov 27 '18 at 23:38






                                • 1




                                  " Is the value of the expression an integer?" If $pi$ is normal then it is an integer. If $pi$ is not normal it is not an integer. The fact that we don't know if it's an integer or not doesn't prevent it from being an integer or a non-integer.
                                  – fleablood
                                  Nov 28 '18 at 0:04










                                • No. A normal number is a real number: en.wikipedia.org/wiki/Normal_number
                                  – marty cohen
                                  Nov 28 '18 at 3:47






                                • 1




                                  Um.... What are you talking about??? Of course normal numbers are real numbers. Everything that isnt complex is a real number. You asked if given the function $f(x) = 1$ if $x$ is normal; $f(x)$ if $x$ is not normal. You asked whether $f(pi)$ is an integer (which it is if it is $1$) or if it was not an integer (which it will not be if it is $frac 12$). And the answer is: We don't know. But EITHER $f(pi)$ is an integer OR it is not. We can NOT say $f(pi)$ is a non-integer, just because we don't know that it is an integer. We don't know it is not.
                                  – fleablood
                                  Nov 28 '18 at 5:38


















                                • I disagree. The number either is or is not an integer whether we know whether it is or is not. So it IS an integer before we simplify it. It's just that before we simplify it we don't KNOW if it is an integer. But that doesn't mean it isn't an integer. (because ... it is.)
                                  – fleablood
                                  Nov 27 '18 at 21:22










                                • Suppose the expression encodes something we don't know the answer for - for example, 1 if $pi$ is a normal number, 1/2 if it is not. Is the value of the expression an integer?
                                  – marty cohen
                                  Nov 27 '18 at 23:38






                                • 1




                                  " Is the value of the expression an integer?" If $pi$ is normal then it is an integer. If $pi$ is not normal it is not an integer. The fact that we don't know if it's an integer or not doesn't prevent it from being an integer or a non-integer.
                                  – fleablood
                                  Nov 28 '18 at 0:04










                                • No. A normal number is a real number: en.wikipedia.org/wiki/Normal_number
                                  – marty cohen
                                  Nov 28 '18 at 3:47






                                • 1




                                  Um.... What are you talking about??? Of course normal numbers are real numbers. Everything that isnt complex is a real number. You asked if given the function $f(x) = 1$ if $x$ is normal; $f(x)$ if $x$ is not normal. You asked whether $f(pi)$ is an integer (which it is if it is $1$) or if it was not an integer (which it will not be if it is $frac 12$). And the answer is: We don't know. But EITHER $f(pi)$ is an integer OR it is not. We can NOT say $f(pi)$ is a non-integer, just because we don't know that it is an integer. We don't know it is not.
                                  – fleablood
                                  Nov 28 '18 at 5:38
















                                I disagree. The number either is or is not an integer whether we know whether it is or is not. So it IS an integer before we simplify it. It's just that before we simplify it we don't KNOW if it is an integer. But that doesn't mean it isn't an integer. (because ... it is.)
                                – fleablood
                                Nov 27 '18 at 21:22




                                I disagree. The number either is or is not an integer whether we know whether it is or is not. So it IS an integer before we simplify it. It's just that before we simplify it we don't KNOW if it is an integer. But that doesn't mean it isn't an integer. (because ... it is.)
                                – fleablood
                                Nov 27 '18 at 21:22












                                Suppose the expression encodes something we don't know the answer for - for example, 1 if $pi$ is a normal number, 1/2 if it is not. Is the value of the expression an integer?
                                – marty cohen
                                Nov 27 '18 at 23:38




                                Suppose the expression encodes something we don't know the answer for - for example, 1 if $pi$ is a normal number, 1/2 if it is not. Is the value of the expression an integer?
                                – marty cohen
                                Nov 27 '18 at 23:38




                                1




                                1




                                " Is the value of the expression an integer?" If $pi$ is normal then it is an integer. If $pi$ is not normal it is not an integer. The fact that we don't know if it's an integer or not doesn't prevent it from being an integer or a non-integer.
                                – fleablood
                                Nov 28 '18 at 0:04




                                " Is the value of the expression an integer?" If $pi$ is normal then it is an integer. If $pi$ is not normal it is not an integer. The fact that we don't know if it's an integer or not doesn't prevent it from being an integer or a non-integer.
                                – fleablood
                                Nov 28 '18 at 0:04












                                No. A normal number is a real number: en.wikipedia.org/wiki/Normal_number
                                – marty cohen
                                Nov 28 '18 at 3:47




                                No. A normal number is a real number: en.wikipedia.org/wiki/Normal_number
                                – marty cohen
                                Nov 28 '18 at 3:47




                                1




                                1




                                Um.... What are you talking about??? Of course normal numbers are real numbers. Everything that isnt complex is a real number. You asked if given the function $f(x) = 1$ if $x$ is normal; $f(x)$ if $x$ is not normal. You asked whether $f(pi)$ is an integer (which it is if it is $1$) or if it was not an integer (which it will not be if it is $frac 12$). And the answer is: We don't know. But EITHER $f(pi)$ is an integer OR it is not. We can NOT say $f(pi)$ is a non-integer, just because we don't know that it is an integer. We don't know it is not.
                                – fleablood
                                Nov 28 '18 at 5:38




                                Um.... What are you talking about??? Of course normal numbers are real numbers. Everything that isnt complex is a real number. You asked if given the function $f(x) = 1$ if $x$ is normal; $f(x)$ if $x$ is not normal. You asked whether $f(pi)$ is an integer (which it is if it is $1$) or if it was not an integer (which it will not be if it is $frac 12$). And the answer is: We don't know. But EITHER $f(pi)$ is an integer OR it is not. We can NOT say $f(pi)$ is a non-integer, just because we don't know that it is an integer. We don't know it is not.
                                – fleablood
                                Nov 28 '18 at 5:38



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