Does the polynomial $P(x,y)=x-5y$ have all real numbers as its roots?
$begingroup$
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
edited Dec 30 '18 at 12:18
Blue
47.7k870151
asked Dec 30 '18 at 11:38
user629353user629353
1147
$endgroup$
|
show 1 more comment
$begingroup$
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
edited Dec 30 '18 at 12:18
Blue
47.7k870151
asked Dec 30 '18 at 11:38
user629353user629353
1147
$endgroup$
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
|
show 1 more comment
$begingroup$
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
edited Dec 30 '18 at 12:18
Blue
47.7k870151
asked Dec 30 '18 at 11:38
user629353user629353
1147
$endgroup$
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
edited Dec 30 '18 at 12:18
Blue
47.7k870151
asked Dec 30 '18 at 11:38
user629353user629353
1147
edited Dec 30 '18 at 12:18
Blue
47.7k870151
asked Dec 30 '18 at 11:38
user629353user629353
1147
edited Dec 30 '18 at 12:18
Blue
47.7k870151
edited Dec 30 '18 at 12:18
Blue
47.7k870151
edited Dec 30 '18 at 12:18
Blue
47.7k870151
47.7k870151
asked Dec 30 '18 at 11:38
user629353user629353
1147
asked Dec 30 '18 at 11:38
user629353user629353
1147
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
|
show 1 more comment
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
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
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 $(x,y,z)$
answered Dec 30 '18 at 12:13
harshit54harshit54
346113
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
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 $(x,y,z)$
answered Dec 30 '18 at 12:13
harshit54harshit54
346113
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $(x,y,z)$
answered Dec 30 '18 at 12:13
harshit54harshit54
346113
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $(x,y,z)$
answered Dec 30 '18 at 12:13
harshit54harshit54
346113
$endgroup$
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 $(x,y,z)$
answered Dec 30 '18 at 12:13
harshit54harshit54
346113
answered Dec 30 '18 at 12:13
harshit54harshit54
346113
answered Dec 30 '18 at 12:13
harshit54harshit54
346113
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
add a comment |
$begingroup$
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
add a comment |
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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