The difference between 'solution' and 'root'
I am wondering about the difference between the following demands:
Prove that P(x) has at least one root.
Prove that P(x) has at least one solution.
Are they the same?
The background to my question:
Let $c \in \Bbb R$. Prove that the equation: $$\frac{1}{\ln x} - \frac{1}{x-1}$$
Has only one solution in $(0,1)$.
Here is what I have in mind:
- Show that if $f'$ has no roots, then $f$ has one root at most.
- Calculate the derivative.
- Show that when $\frac{1}{2}<x<1$, $f'(x)<0$
- Show that $f'(1)=0$. This means that 0 is a local minimum point in $(0,1)$
- Therefore, $f$ has at most one root.
- Show from Intermediate value theorem that $f$ has a root in $(0,1)$.
3 Answers
$\begingroup$Yes, a root is basically the same as a solution. We often use the work "root" when we talk about polynomials.
Now, this doesn't mean that the words are interchangeable. We can say
The roots of the polynomial are ...
We can't say
The solutions to the polynomial are ...
Solutions are always to equations. Roots are for/of polynomials (functions).
You write that the equation $$ \frac{1}{\ln(x)} - \frac{1}{x-1} $$ has a solution, but what you have written isn't an equation because you don't have an equal sign. Instead you probably want to say that the equation $$ \frac{1}{\ln(x)} - \frac{1}{x-1} =0 $$ has a solution ...
Example: Let $f(x) = x^2 -3x + 2$. Then $f$ is a polynomial. We can say the following
$\endgroup$ 2 $\begingroup$$1$ and $2$ are the two roots of the polynomial $f$.
$1$ and $2$ are the two solutions to the equation $f(x) = 0$.
Functions have roots, but equations have solutions.
So, we'd say that
the function $P$ has at least one root,
but not that "$P$ has at least one solution".
Conversely, we'd say that
the equation $P(x) = 0$ has at least one solution
but not that "$P(x) = 0$ has at least one root".
Of course the two notions are related, in that, by definition, $x_0$ is a root of $P$ iff $P(x_0) = 0$, or equivalently, if $x = x_0$ is a solution of $P(x) = 0$.
$\endgroup$ 4 $\begingroup$Prove that P(x) has at least one root.
The solution to the above will involve radicals, like square roots. This means you'll find a zero, but not a solution. You will solve your problem, however.
Here's an example $x^2-x-1=0$
Prove that P(x) has at least one solution.
The solution can take any form, and instead of showing one side of the equation has zeros you show that there are values where both sides of the equation are equal.
For example $e^x=7$
$\endgroup$
