Does the polynomial $P(x,y)=x-5y$ have all real numbers as its roots?












0












$begingroup$



Does the polynomial $P(x,y)=x-5y$ have all real numbers as its roots?




I took the following steps to find its roots:



Convert its output zero, $P(x,y)=0$:



$$x-5y=0 qquadtoqquad x=5y$$



For each value if $x$ in $mathbb{R}$, there is a corresponding value of $y$ we get. Thus, we get infinitely many ordered pair values of $x$ and $y$. And, on putting value of each pair, will make the polynomial $P(x,y)$ zero. Thus, $P(x,y)$ has infinitely many roots.



Is these steps true, if yes then are there other examples of infinite root polynomials?






polynomials







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Yes, it's ok. Remember that the sentence "a polynomial of degree $n$ has $n$ roots" is true in ONE variable (over $mathbb{C}$)
    $endgroup$
    – Martín Vacas Vignolo
    Dec 30 '18 at 11:50










  • $begingroup$
    Is there any example of polynomial in one variable having infinite roots
    $endgroup$
    – user629353
    Dec 30 '18 at 11:54






  • 1




    $begingroup$
    The only one is the polynomial $p(x)=0$
    $endgroup$
    – Martín Vacas Vignolo
    Dec 30 '18 at 11:55










  • $begingroup$
    But zero is neither multivariate nor univariate
    $endgroup$
    – user629353
    Dec 30 '18 at 11:56






  • 1




    $begingroup$
    Saying that all real numbers are it's roots is incorrect. You should say that there as infinite possible ordered pairs $(x,y)$ which satisfy this equation.
    $endgroup$
    – harshit54
    Dec 30 '18 at 11:59
















0












$begingroup$



Does the polynomial $P(x,y)=x-5y$ have all real numbers as its roots?




I took the following steps to find its roots:



Convert its output zero, $P(x,y)=0$:



$$x-5y=0 qquadtoqquad x=5y$$



For each value if $x$ in $mathbb{R}$, there is a corresponding value of $y$ we get. Thus, we get infinitely many ordered pair values of $x$ and $y$. And, on putting value of each pair, will make the polynomial $P(x,y)$ zero. Thus, $P(x,y)$ has infinitely many roots.



Is these steps true, if yes then are there other examples of infinite root polynomials?






polynomials







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Yes, it's ok. Remember that the sentence "a polynomial of degree $n$ has $n$ roots" is true in ONE variable (over $mathbb{C}$)
    $endgroup$
    – Martín Vacas Vignolo
    Dec 30 '18 at 11:50










  • $begingroup$
    Is there any example of polynomial in one variable having infinite roots
    $endgroup$
    – user629353
    Dec 30 '18 at 11:54






  • 1




    $begingroup$
    The only one is the polynomial $p(x)=0$
    $endgroup$
    – Martín Vacas Vignolo
    Dec 30 '18 at 11:55










  • $begingroup$
    But zero is neither multivariate nor univariate
    $endgroup$
    – user629353
    Dec 30 '18 at 11:56






  • 1




    $begingroup$
    Saying that all real numbers are it's roots is incorrect. You should say that there as infinite possible ordered pairs $(x,y)$ which satisfy this equation.
    $endgroup$
    – harshit54
    Dec 30 '18 at 11:59














0












0








0





$begingroup$



Does the polynomial $P(x,y)=x-5y$ have all real numbers as its roots?




I took the following steps to find its roots:



Convert its output zero, $P(x,y)=0$:



$$x-5y=0 qquadtoqquad x=5y$$



For each value if $x$ in $mathbb{R}$, there is a corresponding value of $y$ we get. Thus, we get infinitely many ordered pair values of $x$ and $y$. And, on putting value of each pair, will make the polynomial $P(x,y)$ zero. Thus, $P(x,y)$ has infinitely many roots.



Is these steps true, if yes then are there other examples of infinite root polynomials?






polynomials







share|cite|improve this question











$endgroup$





Does the polynomial $P(x,y)=x-5y$ have all real numbers as its roots?




I took the following steps to find its roots:



Convert its output zero, $P(x,y)=0$:



$$x-5y=0 qquadtoqquad x=5y$$



For each value if $x$ in $mathbb{R}$, there is a corresponding value of $y$ we get. Thus, we get infinitely many ordered pair values of $x$ and $y$. And, on putting value of each pair, will make the polynomial $P(x,y)$ zero. Thus, $P(x,y)$ has infinitely many roots.



Is these steps true, if yes then are there other examples of infinite root polynomials?






polynomials




polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 12:18









Blue

47.7k870151




47.7k870151










asked Dec 30 '18 at 11:38









user629353user629353

1147




1147








  • 2




    $begingroup$
    Yes, it's ok. Remember that the sentence "a polynomial of degree $n$ has $n$ roots" is true in ONE variable (over $mathbb{C}$)
    $endgroup$
    – Martín Vacas Vignolo
    Dec 30 '18 at 11:50










  • $begingroup$
    Is there any example of polynomial in one variable having infinite roots
    $endgroup$
    – user629353
    Dec 30 '18 at 11:54






  • 1




    $begingroup$
    The only one is the polynomial $p(x)=0$
    $endgroup$
    – Martín Vacas Vignolo
    Dec 30 '18 at 11:55










  • $begingroup$
    But zero is neither multivariate nor univariate
    $endgroup$
    – user629353
    Dec 30 '18 at 11:56






  • 1




    $begingroup$
    Saying that all real numbers are it's roots is incorrect. You should say that there as infinite possible ordered pairs $(x,y)$ which satisfy this equation.
    $endgroup$
    – harshit54
    Dec 30 '18 at 11:59














  • 2




    $begingroup$
    Yes, it's ok. Remember that the sentence "a polynomial of degree $n$ has $n$ roots" is true in ONE variable (over $mathbb{C}$)
    $endgroup$
    – Martín Vacas Vignolo
    Dec 30 '18 at 11:50










  • $begingroup$
    Is there any example of polynomial in one variable having infinite roots
    $endgroup$
    – user629353
    Dec 30 '18 at 11:54






  • 1




    $begingroup$
    The only one is the polynomial $p(x)=0$
    $endgroup$
    – Martín Vacas Vignolo
    Dec 30 '18 at 11:55










  • $begingroup$
    But zero is neither multivariate nor univariate
    $endgroup$
    – user629353
    Dec 30 '18 at 11:56






  • 1




    $begingroup$
    Saying that all real numbers are it's roots is incorrect. You should say that there as infinite possible ordered pairs $(x,y)$ which satisfy this equation.
    $endgroup$
    – harshit54
    Dec 30 '18 at 11:59








2




2




$begingroup$
Yes, it's ok. Remember that the sentence "a polynomial of degree $n$ has $n$ roots" is true in ONE variable (over $mathbb{C}$)
$endgroup$
– Martín Vacas Vignolo
Dec 30 '18 at 11:50




$begingroup$
Yes, it's ok. Remember that the sentence "a polynomial of degree $n$ has $n$ roots" is true in ONE variable (over $mathbb{C}$)
$endgroup$
– Martín Vacas Vignolo
Dec 30 '18 at 11:50












$begingroup$
Is there any example of polynomial in one variable having infinite roots
$endgroup$
– user629353
Dec 30 '18 at 11:54




$begingroup$
Is there any example of polynomial in one variable having infinite roots
$endgroup$
– user629353
Dec 30 '18 at 11:54




1




1




$begingroup$
The only one is the polynomial $p(x)=0$
$endgroup$
– Martín Vacas Vignolo
Dec 30 '18 at 11:55




$begingroup$
The only one is the polynomial $p(x)=0$
$endgroup$
– Martín Vacas Vignolo
Dec 30 '18 at 11:55












$begingroup$
But zero is neither multivariate nor univariate
$endgroup$
– user629353
Dec 30 '18 at 11:56




$begingroup$
But zero is neither multivariate nor univariate
$endgroup$
– user629353
Dec 30 '18 at 11:56




1




1




$begingroup$
Saying that all real numbers are it's roots is incorrect. You should say that there as infinite possible ordered pairs $(x,y)$ which satisfy this equation.
$endgroup$
– harshit54
Dec 30 '18 at 11:59




$begingroup$
Saying that all real numbers are it's roots is incorrect. You should say that there as infinite possible ordered pairs $(x,y)$ which satisfy this equation.
$endgroup$
– harshit54
Dec 30 '18 at 11:59










1 Answer
1






active

oldest

votes


















-1












$begingroup$

Depends on the way the zero polynomial is defined. For example



$f(x) = 0$



All the solutions lie on the line $y = 0$. This is satisfied by all the real numbers.



$f(x, y) = 0$



All the solutions lie on the plane $z = 0$. This is satisfied by all the ordered pairs $(x,y)$



$f(x, y, z) = 0$



All the solutions lie on the ?space? $w = 0$. This is satisfied by all the ordered triplets $(x,y,z)$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is it situation based or not defined?
    $endgroup$
    – user629353
    Dec 30 '18 at 12:26












  • $begingroup$
    Well it depends on the number of inputs to the function.
    $endgroup$
    – harshit54
    Dec 30 '18 at 12:27












  • $begingroup$
    Is similar argument to what you are saying written on any webpage or book, if there then please link it to me
    $endgroup$
    – user629353
    Dec 30 '18 at 12:29










  • $begingroup$
    @user629353 Sorry I can't give you any proof to support my claims.
    $endgroup$
    – harshit54
    Dec 30 '18 at 12:38











Your Answer





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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









-1












$begingroup$

Depends on the way the zero polynomial is defined. For example



$f(x) = 0$



All the solutions lie on the line $y = 0$. This is satisfied by all the real numbers.



$f(x, y) = 0$



All the solutions lie on the plane $z = 0$. This is satisfied by all the ordered pairs $(x,y)$



$f(x, y, z) = 0$



All the solutions lie on the ?space? $w = 0$. This is satisfied by all the ordered triplets $(x,y,z)$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is it situation based or not defined?
    $endgroup$
    – user629353
    Dec 30 '18 at 12:26












  • $begingroup$
    Well it depends on the number of inputs to the function.
    $endgroup$
    – harshit54
    Dec 30 '18 at 12:27












  • $begingroup$
    Is similar argument to what you are saying written on any webpage or book, if there then please link it to me
    $endgroup$
    – user629353
    Dec 30 '18 at 12:29










  • $begingroup$
    @user629353 Sorry I can't give you any proof to support my claims.
    $endgroup$
    – harshit54
    Dec 30 '18 at 12:38
















-1












$begingroup$

Depends on the way the zero polynomial is defined. For example



$f(x) = 0$



All the solutions lie on the line $y = 0$. This is satisfied by all the real numbers.



$f(x, y) = 0$



All the solutions lie on the plane $z = 0$. This is satisfied by all the ordered pairs $(x,y)$



$f(x, y, z) = 0$



All the solutions lie on the ?space? $w = 0$. This is satisfied by all the ordered triplets $(x,y,z)$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is it situation based or not defined?
    $endgroup$
    – user629353
    Dec 30 '18 at 12:26












  • $begingroup$
    Well it depends on the number of inputs to the function.
    $endgroup$
    – harshit54
    Dec 30 '18 at 12:27












  • $begingroup$
    Is similar argument to what you are saying written on any webpage or book, if there then please link it to me
    $endgroup$
    – user629353
    Dec 30 '18 at 12:29










  • $begingroup$
    @user629353 Sorry I can't give you any proof to support my claims.
    $endgroup$
    – harshit54
    Dec 30 '18 at 12:38














-1












-1








-1





$begingroup$

Depends on the way the zero polynomial is defined. For example



$f(x) = 0$



All the solutions lie on the line $y = 0$. This is satisfied by all the real numbers.



$f(x, y) = 0$



All the solutions lie on the plane $z = 0$. This is satisfied by all the ordered pairs $(x,y)$



$f(x, y, z) = 0$



All the solutions lie on the ?space? $w = 0$. This is satisfied by all the ordered triplets $(x,y,z)$






share|cite|improve this answer









$endgroup$



Depends on the way the zero polynomial is defined. For example



$f(x) = 0$



All the solutions lie on the line $y = 0$. This is satisfied by all the real numbers.



$f(x, y) = 0$



All the solutions lie on the plane $z = 0$. This is satisfied by all the ordered pairs $(x,y)$



$f(x, y, z) = 0$



All the solutions lie on the ?space? $w = 0$. This is satisfied by all the ordered triplets $(x,y,z)$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 30 '18 at 12:13









harshit54harshit54

346113




346113












  • $begingroup$
    Is it situation based or not defined?
    $endgroup$
    – user629353
    Dec 30 '18 at 12:26












  • $begingroup$
    Well it depends on the number of inputs to the function.
    $endgroup$
    – harshit54
    Dec 30 '18 at 12:27












  • $begingroup$
    Is similar argument to what you are saying written on any webpage or book, if there then please link it to me
    $endgroup$
    – user629353
    Dec 30 '18 at 12:29










  • $begingroup$
    @user629353 Sorry I can't give you any proof to support my claims.
    $endgroup$
    – harshit54
    Dec 30 '18 at 12:38


















  • $begingroup$
    Is it situation based or not defined?
    $endgroup$
    – user629353
    Dec 30 '18 at 12:26












  • $begingroup$
    Well it depends on the number of inputs to the function.
    $endgroup$
    – harshit54
    Dec 30 '18 at 12:27












  • $begingroup$
    Is similar argument to what you are saying written on any webpage or book, if there then please link it to me
    $endgroup$
    – user629353
    Dec 30 '18 at 12:29










  • $begingroup$
    @user629353 Sorry I can't give you any proof to support my claims.
    $endgroup$
    – harshit54
    Dec 30 '18 at 12:38
















$begingroup$
Is it situation based or not defined?
$endgroup$
– user629353
Dec 30 '18 at 12:26






$begingroup$
Is it situation based or not defined?
$endgroup$
– user629353
Dec 30 '18 at 12:26














$begingroup$
Well it depends on the number of inputs to the function.
$endgroup$
– harshit54
Dec 30 '18 at 12:27






$begingroup$
Well it depends on the number of inputs to the function.
$endgroup$
– harshit54
Dec 30 '18 at 12:27














$begingroup$
Is similar argument to what you are saying written on any webpage or book, if there then please link it to me
$endgroup$
– user629353
Dec 30 '18 at 12:29




$begingroup$
Is similar argument to what you are saying written on any webpage or book, if there then please link it to me
$endgroup$
– user629353
Dec 30 '18 at 12:29












$begingroup$
@user629353 Sorry I can't give you any proof to support my claims.
$endgroup$
– harshit54
Dec 30 '18 at 12:38




$begingroup$
@user629353 Sorry I can't give you any proof to support my claims.
$endgroup$
– harshit54
Dec 30 '18 at 12:38


















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