I have seen questions of this type being solved as follows : $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$'s value does not change if we add an $x$ to the expression and square root it. Let the value of this expression be $y$. So $$\sqrt{x+y} = y \implies x+y = y^2 \implies y^2-y-x=0$$ Using the quadratic formula, we obtain the value of $y$ as : $$\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(-x)}}{2(1)} = \dfrac{1\pm\sqrt{1+4x}}{2} = \sqrt{x+\sqrt{x+\sqrt{x+...}}}$$ This has been used to solve $\sqrt{6+\sqrt{6+\sqrt{6+...}}}$ in my Mathematics textbook. Now, this method would work perfectly, assuming that a value for $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ exists. If a value for this expression does not exist, this would be similar to Numberphile's popular $\displaystyle\sum_{n=1}^\infty n = \dfrac{-1}{12}$ which is doubtlessly wrong. So, does a value for $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ exist? Why/why not? Thanks!
3167
(Fitzpatrick Advanced Calculus 2e, Sec. 2.4 #12)  For $c \gt 0$, consider the quadratic equation $x^2 - x - c = 0, x &gt; 0$.  Define the sequence $\{x_n\}$ recursively by fixing $|x_1| \lt c$ and then, if $n$ is an index for which $x_n$ has been defined, defining  $$x_{n+1} = \sqrt{c+x_n}$$  Prove that the sequence $\{x_n\}$ converges monotonically to the solution of the above equation.  Note: The answers below might assume $x_1 \gt 0$, but they still work, as we have $x_3 \gt 0$.    This is being repurposed in an effort to cut down on duplicates, see here: .  and here: .
21646
Prove that $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational   My attempt:-     Suppose $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is rational, then for some $x\in\mathbb{Q}$ we have $$\sqrt{6}-\sqrt{2}-\sqrt{3}=x$$ Rewriting this equation as $$\sqrt{6}-x=\sqrt{2}+\sqrt{3}$$ and now squaring this we get $$ 6-2x\sqrt{6}+x^2=5+2\sqrt{6}$$. This implies that $$\sqrt{6}=\frac{x^2-1}{2+2x}$$ but this is absurd as RHS of the above equation is rational but we know that $\sqrt6$ is irrational. Therefore , $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational. Does this look good? Have I written it properly? Is there any other proof besides this..like one using geometry? Thank you.
91275
If :- $$a^2x^3 + b^2y^3 + c^2z^3 = p^5$$ $$ax^2 = by^2 = cz^2$$ $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{p}$$ Find $\sqrt a  + \sqrt b + \sqrt c$ in terms of $p$ .  What I Tried :- How do you use the information here? I assumed that :- $$ax^2 = by^2 = cz^2 = k$$ And then I can write the $1$st equation as :- $$k(ax + by + cz) = p^5$$ But I guess that just bring another extra variable into the scene so it does not help . How can I use the $3$rd equation? I can write it as $\frac{xyz}{xy + yz + zx} = p$ . How do you use it?
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Show that if $x_n \to x$ then $\sqrt{x_n} \to \sqrt{x}$ I have been stuck on this for a while. I tried $$|\sqrt{x_n} - \sqrt{x}| = \frac{|x_n - x|}{|\sqrt{x_n} + \sqrt{x}|},$$ and then I at least can get the top to be as small as I want, so I have $$\frac{\epsilon}{|\sqrt{x+\epsilon} + \sqrt{x}|},$$ but I get stuck here at choosing the N, and I don't know if my first step in breaking down the absolute value is legitimate. Please help.
122351
Find the maximum value of the expression :   $${\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}$$   where $x,y,z$ are real numbers satisfying the condition that $x+y+z=1$.   Taking $x=y=z=\frac {1}{3}$, I get the expression as $\frac {3x}{1+x^2}$, which is equal to $\frac {1}{1+{\frac{1}{9}}}$ or $\frac {9}{10}$.  How can I actually solve the problem without making unnecessary assumptions ?
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Can the expression $\sqrt{n} + \sqrt{m}$ be rational if neither $n,m \in \mathbb{N}$ are perfect squares? It doesn't seem likely, the only way that could happen is if for example $\sqrt{m} = a-\sqrt{n}, \ \ a \in \mathbb{Q}$, which I don't think is possible, but how to show it?
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I am trying to help my daughter on her math homework and I am having some trouble on some equation solving steps. My current major concern relies on understanding why $\sqrt{8}/2$ equal to $\sqrt{2}$.  Thanks in advance
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As far as I see, $\sqrt{x^2}$ is not $x$, but $|x|$, meaning the "absolute". I totally get this, because $x^2$ is positive, if $x$ is negative, so $\sqrt{y}$, whether $y = 10^2$ or $y = -10^2$: $y$ is positive.  But then I remember that $\sqrt(x)$ is the same as $x^{1/2}$ and thus, $\sqrt{x^2}$ is the same as $x^{1}$.   So, as far as I get it, $\sqrt{x^2} = x^{2/2}$.  But then, I can cancel $\frac{2}{2}$ to $1$. So: $\sqrt{x^2} = x^1 = x$.  Where does the absolute value in this derivation of the calculation come from?  I get it why its there: Because $(-x)^2 = x^2$, so the radicand is always positive. But from $\sqrt{x^n} = x^{n/2}$ I dont understand how it gets there.
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My proof: By the way of contradiction, suppose that we found $(x,y) \in \mathbb N^*$ satisfies:$$\frac{x+y}{xy}=5$$ So : $$x+y=5xy \\ \iff x=5xy-y \iff x=y(5x-1)$$ Since $x\in \mathbb N^* $ so the minimum value of $5x$ is $5&gt;1 \implies y\mid x$ By the same method we have: $$y=x(5y-1) \implies x\mid y$$ $x\mid y, y\mid x \implies x=y$, Now we can write the original equation as : $$\frac{2x}{x^2}=5 \iff 2=5x \implies x\notin \mathbb N^* $$ Therefore there is no solution to this equation: $$\frac{x+y}{xy} =5$$ Does this considered a proof?
65035
I need help proving the following statement:  For all integers $x$ and $y$, if $x^3+x = y^3+y$ then $x = y$  The statement is true, I just need to know the thought process, or a lead in the right direction. I think I might have to use a contradiction, but I don't know where to begin.  Any help would be much appreciated.
28857
I want to solve this system of equations, I have been out of Maths for a long!!  $$\sqrt{x}+y=7$$  $$\sqrt{y}+x=11$$  Just wondering easiest step to find values for $x$ and $y$ from the above equations?
71323
Prove: $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$$ for all x, y positive   $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}-\sqrt{x}-\sqrt{y}\geq 0$$  $$\frac{x\sqrt{x}+y\sqrt{y}-x\sqrt{y}-y\sqrt{x}}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{x(\sqrt{x}-\sqrt{y})+y(\sqrt{y}-\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{x(\sqrt{x}-\sqrt{y})-y(-\sqrt{y}+\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{(x-y)(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{(\sqrt{x}-\sqrt{y})^2(\sqrt{x}+\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$  All the elements are positive and if $\sqrt{x}=\sqrt{y}$ we get $0$  Is the proof valid?
88475
let $ y=\displaystyle \sqrt{x+\sqrt{{x}+\sqrt{x+\cdots}}}$, i'm really interesting to know how do I find :  $\displaystyle \frac{dy}{dx}$ ?.  Note: I have used the definition of derivative of the square root function but i don't succed .  Thank you for any help
21368
You learn in algebra that  $$\sqrt{ab}=\sqrt{a} \sqrt{b}$$ and that  $$\sqrt{\frac ab}=\frac {\sqrt a}{\sqrt b}$$ You also learn to never make the fatal mistake of thinking $$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$$  However, I am wondering if there is a rule for $\sqrt{a+b}$. I would think it would be pretty complex, if it exists at all.  $$\sqrt{a+b}=\text{?}$$
44571
In my answer to the recent question , @GEdgar correctly raised the issue that the proof is incomplete unless I show that the intermediate expressions do converge to a (finite) limit. One such quantity was the nested radical  $$ \sqrt{1 + \sqrt{1+\sqrt{1 + \sqrt{1 + \cdots}}}} \tag{1} $$  To assign a value $Y$ to such an expression, I proposed the following definition. Define the sequence $\{ y_n \}$ by:  $$ y_1 = \sqrt{1}, y_{n+1} = \sqrt{1+y_n}. $$ Then we say that this expression evaluates to $Y$ if the sequence $y_n$ converges to $Y$.   For the expression (1), I could show that the $y_n$ converges to $\phi = (\sqrt{5}+1)/2$. (To give more details, I showed, by induction, that $y_n$ increases monotonically and is bounded by $\phi$, so that it has a limit $Y &lt; \infty$. Furthermore, this limit must satisfy $Y = \sqrt{1+Y}$.) Hence we could safely say (1) evaluates to $\phi$, and all seems to be good.  My trouble. Let us now test my proposed idea with a more general expression of the form $$\sqrt{a_1 + \sqrt{a_2 + \sqrt{a_3 + \sqrt{a_4+\cdots}}}} \tag{2}$$  (Note that the linked question involves one such expression, with $a_n = 5^{2^n}$.) How do we decide if this expression converges? Mimicking the above definition, we can write: $$ y_1 = \sqrt{a_1}, y_{n+1} = \sqrt{a_{n+1}+y_n}. $$ However, unrolling this definition, one get the sequence $$ \sqrt{a_1}, \sqrt{a_{2}+ \sqrt{a_1}}, \sqrt{a_3 + \sqrt{a_2 + \sqrt{a_1}}}, \sqrt{a_4+\sqrt{a_3 + \sqrt{a_2 + \sqrt{a_1}}}}, \ldots $$ but this seems little to do with the expression (2) that we started with.   I could not come up with any satisfactory ways to resolve the issue. So, my question is:     How do I rigorously define when an expression of the form (2) converges, and also assign a value to it when it does converge?   Thanks.
114141
Q1 - Why is $\sqrt{-x}*\sqrt{-x}=-x?$  Q2 - I was thinking it would be:  $\sqrt{-x}*\sqrt{-x}=\sqrt{-x*-x}=\sqrt{x^2}$ but apparently not (why not?)  Q3 - What are the formal algebra rules to use? Can I calculate this without using i such as in: $\sqrt{-x}*\sqrt{-x}=i\sqrt{x}*i\sqrt{x}=-\sqrt{x^2}=-x$.
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Prove by contradiction (not using a calculator) that $\sqrt6 + \sqrt2 &lt; \sqrt{15}$.   How do you approach such a problem?  I need to admit that I'm completely new to proof writing and I have completely no experience in answering that kind of questions. I tried to square it as I read online but I'm stuck with having $\sqrt3$ and not knowing how to get rid of it to have the answer as clear as the sun. squaring them yielded: $(\sqrt2 + \sqrt6)^2 =  8 + 4\sqrt3 &lt; 15$ which indeed holds but the problem is that I am not allowed to use calculator to check the little difference.  My final answer was $4\sqrt3 &lt; 7$.