How to prove that $\sqrt{2+\sqrt{2+\sqrt{2+\dots}} }=2$
120686
Consider the sequence defined by $a_1 = \sqrt{2}$, $a_2 = \sqrt{2 + \sqrt{2}}$, so that in general, $a_n = \sqrt{2 + a_{n - 1}}$ for $n &gt; 1$. I know 2 is an upper bound of this sequence (I proved this by induction). Is there a way to show that this sequence converges to 2? What I think is that the key step is to prove 2 is the least upper bound of this sequence. But how?
21748
I am having issues solving these problems: I tried using $(|a|+|b|)^2 \geq 0$ and $(|a|-|b|)^2 \geq 0$ but I am having problems constructing the proof. I need to gain some intuition on how to proceed.     Show that for $a,b\in (-\infty,+\infty)$,   the following inequalities apply:   a) $2|ab| \leq a^2 + b^2$  b) $|a|+|b| \leq \sqrt {2}(a^2+b^2)^{1/2}$
87058
This is from Spivak.  Prove that $\sqrt{2}+\sqrt{3}$ is irrational.  So far, I have this:  If $\sqrt{2}+\sqrt{3}$ is rational, then it can be written as $\frac{p}{q}$ with integral $p, q$ and in lowest terms.   $$\sqrt{2}+\sqrt{3}=\frac{p}{q}$$ $$2\sqrt{6}+5 =\frac{p^2}{q^2}$$ $$(2\sqrt{6}+5)q^2=p^2$$  And that's about where I get stuck. In a similar question (prove that $\sqrt{2}+\sqrt{6}$ is irrational,) I was able to show that both $p,q$ had to be even which is impossible. I obviously can't apply this trick here. Any hints?
41373
Let $x,y,z\ge 0$,and $x+y+z=2$, show that $$\sqrt{1-xy}+\sqrt{1-yz}+\sqrt{1-xz}\ge 2$$  Mt try: $$\Longleftrightarrow 3-(xy+yz+zx)+2\sum_\text{cyc}\sqrt{(1-xy)(1-yz)}\ge 4$$ or $$\sum_\text{cyc}\sqrt{(1-xy)(1-yz)}\ge\dfrac{1}{2}(1+xy+yz+zx)$$
5287
Let $x_{n} = \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}}$  a) Show that $x_{n} &lt; x_{n+1}$  b) Show that  $x_{n+1}^{2} \leq 1+ \sqrt{2} x_{n}$  Hint : Square $x_{n+1}$ and factor a 2 out of the square root  c) Hence Show that  $x_{n}$ is bounded above by 2. Deduce that $\lim\limits_{n\to \infty} x_{n}$ exists.  Any help? I don't know where to start.
101806
If we assume that: $0\le x_1,x_2,\ldots,x_{10}\le\frac{\pi}{2} $ such that:  $$\sin^2 (x_1) +\sin^2 (x_2)+\cdots+\sin^2(x_{10})=1$$  How to prove that:  $$\frac{\cos(x_1) +\cos(x_2) +\cdots+\cos(x_{10})}{\sin(x_1) +\sin(x_2) +\cdots+\sin(x_{10})} \ge 3$$
576
Here is a statement that seems prima facie obvious, but when I try to prove it, I am lost.  Let $x_1 , x_2, \dots, x_k$ be complex numbers satisfying:  $$x_1 + x_2+ \dots + x_k = 0$$  $$x_1^2 + x_2^2+ \dots + x_k^2 = 0$$  $$x_1^3 + x_2^3+ \dots + x_k^3 = 0$$  $$\dots$$  Then $x_1 = x_2 = \dots = x_k = 0$.    The statement seems obvious because we have more than $k$ constraints (constraints that are in some sense, "independent") on $k$ variables, so they should determine the variables uniquely.  But my attempts so far of formalizing this intuition have failed.  So, how do you prove this statement?  Is there a generalization of my intuition?
103410
why $1+\frac{1}{1+\frac{1}{\frac{1}{1+\dots}}}$ cannot be equal to $\frac{1-\sqrt{5}}{2}$  I know that some of you will answer because it is positive.I know it but I don't know how to prove it.Maybe the second part is positive how do you know that?Thanks in advance.
86572
How should I approach obvious proofs like these    I have been trying but couldn't find an elegant way to work these. Any help is highly appreciated ! Especially looking for a nice proof/hint for first problem
68476
Here's how it is.  So I was studying some notes on Discrete Structures then I discovered this. Prove that $2^N = \binom{N}{0} + \binom{N}{1} + \binom{N}{2} + \dots + \binom{N}{N}$ Since I was new to this, I didn't quite understand it.  I tried using the binomial expansion but it confused me a little.  Can someone please help me solve this.  That would be great.
118198
I want to prove that $\mathbb{Q}(\sqrt p_{1}+\sqrt p_{2})=\mathbb{Q}(\sqrt p_{1},\sqrt p_{2})$ for $p_{1},p_{2}$ primes. I know this was proved before for the generalized case of $p_{1},....,p_{n}$ primes here:    And every proof of this kind of equality involves Galois theory but I still don't know Galois theory and I'm supposed to prove this with only basic  Field theory(field extensions, irreducible polynomials, algebraic extensions, etc).  By definition $\mathbb{Q}(\sqrt p_{1}+\sqrt p_{2})$ is the smallest field containing $\mathbb{Q}$ and $\sqrt p_{1}+\sqrt p_{2}$, also $\mathbb{Q}(\sqrt p_{1},\sqrt p_{2})$ is the smallest field containing $\sqrt p_{1}$ and $\sqrt p_{2}$. Or  $$\mathbb{Q}(\sqrt{p_{1}},\sqrt{p_{2}})=\{a+b\sqrt{p_{1}}+c\sqrt{p_{2}}+d\sqrt{p_{1}p_{2}} \mid a,b,c,d\in\mathbb{Q}\}$$   $$\mathbb{Q}(\sqrt{p_{1}}+\sqrt{p_{2}}) = \lbrace a+b(\sqrt{p_{1}}+\sqrt{p_{2}}) \mid a,b \in \mathbb{Q} \rbrace  $$  Also I know that $[\mathbb{Q}(\sqrt p_{1},\sqrt p_{2}):\mathbb{Q}]=4$ from this:    Still don't know how to proceed proving this two extensions are the same.
122526
Prove that if n is the product of two numbers that can be written as the sum of two squares then n can be written as the sum of two squares.
116810
How to show  $x_1,x_2, \dots ,x_n \geq 0 $ and $ x_1 + x_2 + \dots + x_n \leq \frac{1}{2} \implies (1-x_1)(1-x_2) \cdots (1-x_n) \geq \frac{1}{2}$
116030
I know how to prove this using an element and subset proof but that's not what I'm being asked to do. I'm suppose to prove it using the distributive law and the associative law but I'm just not seeing how to do this. Can you give me some advice on how to start this. I figure the first thing would be $$((A \cup B) \cap A) \cup ((A \cup B) \cap B^c))$$ But I'm not sure where to go from there.
109854
I am stuck at   $$(a+d)^2 - 4(ad-bc) &lt; 0$$ $$\implies (a+d)^2&lt;4(ad-bc)$$ $$\implies (a+d)&lt;2\sqrt{ad-bc}$$  where $a,b,c,d \ge 0$.  Is there a contradiction to derive here?   Also, the square root on the R.H.S. may not be legit, since I don't have any more information about the relation of $ad$ and $bc$ -- so let's say I took the complex square root, to be safe.    (This lack of information also makes it hard to use the AM-GM inequality.)  Any help is appreciated.  Thanks,
106159
How can I prove that  $$\lim_{a \to 0^{+}} \int_{0}^{a} \frac{1}{\sqrt{\cos(x)-\cos(a)}} \;dx=\frac{\pi}{\sqrt{2}}$$
100164
How to find the real values of $x$ such that : $$x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$