Bolzano's theorem for real closed field.












1












$begingroup$



Take any real closed field $K$ and denote its unique ordering by $leq$. Prove if $f in K[X]$ and $a,b in K$ such that $a<b$ and $f(a) < 0 < f(b)$, then there is $c$ that $f(c)=0$.




I want use the fact that if we have real closed field $K$, so every polynomial $ f in K[X]$ splits into irreducible factors of the form $X-a$ or $(X-a)^2 + b^2$ where $a,b in K[X]$.



Let $f(x) = (x-d)((x-d)^2 + e^2))$ $ $ $d,e in K$ and assume that for some $a,b in K$: $a<b$, $f(a)f(b) < 0. $



$h(x)=(x-d)$ and $g(x)=(x-d)^2+e^2$



$sign(f)=sign(h)sing(g)$. Since $sign(g)=1$, so $sign(f)=sign(h)$. Assume that $sign(h(a))=sign(h(b)).$ It implies that $sign(f(a))=sign(f(b))$ which contradicts with $f(a)f(b)<0$, so we have $h(a)h(b)<0$.



Then $$h(a)<0<h(b)$$
$$a-d <0 <b-d$$
$$c:=d in [a,b]$$ is a root of $f(x)$.



Is that correct?






field-theory ordered-fields







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  • 1




    $begingroup$
    This is correct but you still need to treat the general case when $f$ is a product of possibly more than two irreducible polynomials.
    $endgroup$
    – nombre
    Dec 12 '18 at 20:30
















1












$begingroup$



Take any real closed field $K$ and denote its unique ordering by $leq$. Prove if $f in K[X]$ and $a,b in K$ such that $a<b$ and $f(a) < 0 < f(b)$, then there is $c$ that $f(c)=0$.




I want use the fact that if we have real closed field $K$, so every polynomial $ f in K[X]$ splits into irreducible factors of the form $X-a$ or $(X-a)^2 + b^2$ where $a,b in K[X]$.



Let $f(x) = (x-d)((x-d)^2 + e^2))$ $ $ $d,e in K$ and assume that for some $a,b in K$: $a<b$, $f(a)f(b) < 0. $



$h(x)=(x-d)$ and $g(x)=(x-d)^2+e^2$



$sign(f)=sign(h)sing(g)$. Since $sign(g)=1$, so $sign(f)=sign(h)$. Assume that $sign(h(a))=sign(h(b)).$ It implies that $sign(f(a))=sign(f(b))$ which contradicts with $f(a)f(b)<0$, so we have $h(a)h(b)<0$.



Then $$h(a)<0<h(b)$$
$$a-d <0 <b-d$$
$$c:=d in [a,b]$$ is a root of $f(x)$.



Is that correct?






field-theory ordered-fields







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    This is correct but you still need to treat the general case when $f$ is a product of possibly more than two irreducible polynomials.
    $endgroup$
    – nombre
    Dec 12 '18 at 20:30














1












1








1





$begingroup$



Take any real closed field $K$ and denote its unique ordering by $leq$. Prove if $f in K[X]$ and $a,b in K$ such that $a<b$ and $f(a) < 0 < f(b)$, then there is $c$ that $f(c)=0$.




I want use the fact that if we have real closed field $K$, so every polynomial $ f in K[X]$ splits into irreducible factors of the form $X-a$ or $(X-a)^2 + b^2$ where $a,b in K[X]$.



Let $f(x) = (x-d)((x-d)^2 + e^2))$ $ $ $d,e in K$ and assume that for some $a,b in K$: $a<b$, $f(a)f(b) < 0. $



$h(x)=(x-d)$ and $g(x)=(x-d)^2+e^2$



$sign(f)=sign(h)sing(g)$. Since $sign(g)=1$, so $sign(f)=sign(h)$. Assume that $sign(h(a))=sign(h(b)).$ It implies that $sign(f(a))=sign(f(b))$ which contradicts with $f(a)f(b)<0$, so we have $h(a)h(b)<0$.



Then $$h(a)<0<h(b)$$
$$a-d <0 <b-d$$
$$c:=d in [a,b]$$ is a root of $f(x)$.



Is that correct?






field-theory ordered-fields







share|cite|improve this question









$endgroup$





Take any real closed field $K$ and denote its unique ordering by $leq$. Prove if $f in K[X]$ and $a,b in K$ such that $a<b$ and $f(a) < 0 < f(b)$, then there is $c$ that $f(c)=0$.




I want use the fact that if we have real closed field $K$, so every polynomial $ f in K[X]$ splits into irreducible factors of the form $X-a$ or $(X-a)^2 + b^2$ where $a,b in K[X]$.



Let $f(x) = (x-d)((x-d)^2 + e^2))$ $ $ $d,e in K$ and assume that for some $a,b in K$: $a<b$, $f(a)f(b) < 0. $



$h(x)=(x-d)$ and $g(x)=(x-d)^2+e^2$



$sign(f)=sign(h)sing(g)$. Since $sign(g)=1$, so $sign(f)=sign(h)$. Assume that $sign(h(a))=sign(h(b)).$ It implies that $sign(f(a))=sign(f(b))$ which contradicts with $f(a)f(b)<0$, so we have $h(a)h(b)<0$.



Then $$h(a)<0<h(b)$$
$$a-d <0 <b-d$$
$$c:=d in [a,b]$$ is a root of $f(x)$.



Is that correct?






field-theory ordered-fields




field-theory ordered-fields






share|cite|improve this question













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asked Dec 12 '18 at 19:29









HikiciankaHikicianka

1448




1448








  • 1




    $begingroup$
    This is correct but you still need to treat the general case when $f$ is a product of possibly more than two irreducible polynomials.
    $endgroup$
    – nombre
    Dec 12 '18 at 20:30














  • 1




    $begingroup$
    This is correct but you still need to treat the general case when $f$ is a product of possibly more than two irreducible polynomials.
    $endgroup$
    – nombre
    Dec 12 '18 at 20:30








1




1




$begingroup$
This is correct but you still need to treat the general case when $f$ is a product of possibly more than two irreducible polynomials.
$endgroup$
– nombre
Dec 12 '18 at 20:30




$begingroup$
This is correct but you still need to treat the general case when $f$ is a product of possibly more than two irreducible polynomials.
$endgroup$
– nombre
Dec 12 '18 at 20:30










1 Answer
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$begingroup$

Yes. You assume that
$sign(h(a)) = sign(h(b))$. Then we have $sign(f(a)) = sign(f(b)) $, but we know that $ f(a) < f(b) $. Hence $ h(a)h(b) < 0 $ and we have



$$
left.
begin{array}{l}
h(a) =& a - d\
h(b) =& b - d
end{array}
right}
Longrightarrow a - d < b - d
$$
Let $c := d$, then
$$
a < c < b quad Longrightarrow quad h(a) < h(c) = 0 < h(b)
$$






share|cite|improve this answer









$endgroup$













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    $begingroup$

    Yes. You assume that
    $sign(h(a)) = sign(h(b))$. Then we have $sign(f(a)) = sign(f(b)) $, but we know that $ f(a) < f(b) $. Hence $ h(a)h(b) < 0 $ and we have



    $$
    left.
    begin{array}{l}
    h(a) =& a - d\
    h(b) =& b - d
    end{array}
    right}
    Longrightarrow a - d < b - d
    $$
    Let $c := d$, then
    $$
    a < c < b quad Longrightarrow quad h(a) < h(c) = 0 < h(b)
    $$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Yes. You assume that
      $sign(h(a)) = sign(h(b))$. Then we have $sign(f(a)) = sign(f(b)) $, but we know that $ f(a) < f(b) $. Hence $ h(a)h(b) < 0 $ and we have



      $$
      left.
      begin{array}{l}
      h(a) =& a - d\
      h(b) =& b - d
      end{array}
      right}
      Longrightarrow a - d < b - d
      $$
      Let $c := d$, then
      $$
      a < c < b quad Longrightarrow quad h(a) < h(c) = 0 < h(b)
      $$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Yes. You assume that
        $sign(h(a)) = sign(h(b))$. Then we have $sign(f(a)) = sign(f(b)) $, but we know that $ f(a) < f(b) $. Hence $ h(a)h(b) < 0 $ and we have



        $$
        left.
        begin{array}{l}
        h(a) =& a - d\
        h(b) =& b - d
        end{array}
        right}
        Longrightarrow a - d < b - d
        $$
        Let $c := d$, then
        $$
        a < c < b quad Longrightarrow quad h(a) < h(c) = 0 < h(b)
        $$






        share|cite|improve this answer









        $endgroup$



        Yes. You assume that
        $sign(h(a)) = sign(h(b))$. Then we have $sign(f(a)) = sign(f(b)) $, but we know that $ f(a) < f(b) $. Hence $ h(a)h(b) < 0 $ and we have



        $$
        left.
        begin{array}{l}
        h(a) =& a - d\
        h(b) =& b - d
        end{array}
        right}
        Longrightarrow a - d < b - d
        $$
        Let $c := d$, then
        $$
        a < c < b quad Longrightarrow quad h(a) < h(c) = 0 < h(b)
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '18 at 20:24









        nowek7nowek7

        383




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