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I'm reading a book that says the following (translated): "Before we proceed, let us remember that, given a real number $a > 0$ and a integer $q > 0$, the symbol $\sqrt[q]{a}$ represents a positive real number such that its $q$-power equals to $a$, that is to say it's the only positive solution of $x^q-a=0$".

My whole problem is to show that there is one, and only one, positive real root to $x^q-a=0$. In the case $x^2 - a = 0$ I already couldn't go further. Any ideas? Thanks in advance.

Chickenmancer
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Feripinho
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    For all $q$ the function is monotone increasing for positive $x$. (and it is strictly negative at $x=0$). – lulu Aug 19 '22 at 21:03
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    What level are you at? What class is this for? For $q\in \mathbb Z; q \ge 1$ can you prove that for positive $x, y$ that $x< y \iff x^q < y^q$. Have you defined that the reals have the least upper bound problem. Can you prove that ${w\in \mathbb R| w^q < a}$ is bounded above. Con you prove that if $\alpha = \sup {w|w^q < a}$ then $\alpha^2 \ne a$ both lead to contradictions? [.... are am I assuming an analysis level solution for a calculus or high school algebra level problem?] – fleablood Aug 19 '22 at 21:14
  • There are complex solutions with positive real and imaginary parts – Тyma Gaidash Aug 19 '22 at 21:21
  • There aren't any other solutions because if $0 < x < y$ then $x^q < y^q$ (can you prove that?). So if $x\ne w$ we can not have $x^q =w^q=a$ so $x^q = a$ has at most one solution. As $f(x) = x^q$ is continuous (it is? why? what does that mean?) and we can find $w$ so that $w^q < a$ (can we?) and $z^q > a$ (can we?) that means there must be some value in between $w$ and $z$ that when $q$ed is exactly $a$. (Do you know why?) – fleablood Aug 19 '22 at 21:25
  • @TymaGaidash "There are complex solutions with positive real and imaginary parts " That's true but it is not at all relevant to the question which is about positve roots and there always being one unique positive root. – fleablood Aug 19 '22 at 21:26
  • Further to fleablood's comment/questions, see Rudin PMA Chapter $1$. – Adam Rubinson Aug 19 '22 at 21:39

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Consider the graph of $y=x^q - a$. Then the gradient is $\frac{dy}{dx} = qx^{q-1}$. When $x>0$, since $q>0$, $\frac{dy}{dx} > 0$. In other words, the function is increasing for $x>0$. At $x=0$, $y=-a$ and so $y<0$ since $a>0$. So there must be a positive solution, since the function is negative at 0, and is increasing, continuous and clearly has no upper limit. There must be only one positive solution because the function is increasing.

S34NM68
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Suppose there exists another value, $b$ so that $b^q -a =0,$ and $b\in \Bbb{R}_{>0}.$ We know that $b\ne \sqrt[q]{a},$ so, $b<\sqrt[q]{a}$ or $b>\sqrt[q]{a}.$

Let's suppose that $\sqrt[q]{a}>1,$ and proceed as follows:

First, suppose $b> \sqrt[q]{a},$ then $b^q > a,$ and so $b^q-a > 0,$ but this is not possible, as $b^q-a =0.$

Finally, suppose $b< \sqrt[q]{a},$ then $b^q <a,$ and so $b^q-a <0,$ again this is not possible, as $b^q-a =0.$

Conclude that $\sqrt[q]{a}$ is the only positive solution.

Please attempt the argument where $\sqrt[q]{a}<1,$ using this as a template.

Chickenmancer
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  • This needs to explain why $b > \sqrt[q]a \implies b^2 > a$. Also this doesn't explain what the number $\sqrt[q] a$ is and whether it actually exists at all. – fleablood Aug 19 '22 at 21:28
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Suppose that we know there is a positive number $\sqrt[q]{a}$ such that $(\sqrt[q]{a})^q-a=0$, but we don't know yet if there is a different positive number $x$ such that $x^q-a=0$. Using the identity $$ w^n-z^n=(w-z)(w^{n-1}+w^{n-2}z+\dots+z^{n-1}) $$ we see that the equation $x^q-a=0$ is equivalent to $$ \left(x-\sqrt[q]{a}\right)\left(x^{q-1}+x^{q-2}(\sqrt[q]{a})+\dots+(\sqrt{a})^{q-1}\right)=0 \, . $$ One of the bracketed terms on the LHS must be equal to zero for the above equality to hold. Since the second bracketed expression is positive, we must have $x-\sqrt[q]{a}=0$, and so $x=\sqrt[q]{a}$.

Joe
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