What is the most easy way I can prove $\sqrt 8/2$ is equal to $\sqrt 2$  I have done $(\sqrt 8/2)^2$ but at the end it gives me $\pm \sqrt 2$ and not positive $\sqrt 2$ So how?
120629
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
47577
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 $4\sqrt3 &lt; 7$.
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?
100482
I've often heard that instead of adding up to a little less than one, 1/2 + 1/4 + 1/8... = 1. Is there any way to prove this using equations without using Sigma, or is it just an accepted fact? I need it without Sigma so I can explain it to my little sister.  It is not a duplicate because this one does not use Sigma, and the one marked as duplicate does. I want it to use variables and equations.
70693
How to compute $$\sqrt{(1\sqrt{(2\sqrt{(3\dots)})})}$$ &amp; $$\sqrt{(1+\sqrt{(2+\sqrt{(3+\cdots)})})}$$?  I understand that  $$\sqrt{(1\sqrt{(2\sqrt{(3\dots)})})}=(1^{1/2})(2^{1/4})(3^{1/8})\cdots$$  and  $$\sqrt{(1+\sqrt{(2+\sqrt{(3+\cdots)})})}=(1+(2+(3+(\cdots))^{1/8})^{1/4})^{1/2} $$  How to Proceed further?
124253
Let $a$ and $b$ be positive integers. Prove that: If $a^2$ divides $b^2$, then $a$ divides $b$.   Context: the lecturer wrote this up in my notes without proving it, but I can't seem to figure out why it's true. Would appreciate a solution.
123218
I am having a problem understanding this proof.   Suppose that $\sqrt{2}$ was rational.   Then there would exist positive integers $a,b$ such that $\sqrt{2}=\frac{a}{b}$. Consequently, the set $S=\{k\sqrt{2} : k \text{ and } k\sqrt{2}$ are positive integers $\}$ is nonempty set of positive integers. Therefore, by the well-ordering property, $S$ has a smallest element, say, $s = t \sqrt{2}$.   We have  $s \sqrt{2}-s = s\sqrt{2}-t\sqrt{2}=(s-t)\sqrt{2}$.  Because $s\sqrt{2} = 2t$ and $s$ are both integers,  $s\sqrt{2} - s = s\sqrt{2}-t\sqrt{2} = (s-t)\sqrt{2}$  must also be an integer.   Furthermore, it is positive, because $s\sqrt{2}-s = s(\sqrt{2}-1)$ and $\sqrt{2} &gt; 1$.   It is less than $s$, because $\sqrt{2} &lt; 2$ so that $\sqrt{2}-1 &lt; 1$. This contradicts the choice of $s$ as the smallest positive integer in $S$. It follows that $\sqrt{2}$ is irrational.  My main problem here is how they got the set $S$. My second confusion is the algebra part $s \sqrt{2}-s$. How did they come up with this?   If someone can explain this proof to me it would be great! thank you
91742
I understand that $\sqrt{x^2} = |x|$ because the principal square root is positive.  But since $\sqrt x = x^{\frac{1}{2}}$ shouldn't $\sqrt{x^2} = (x^2)^{\frac{1}{2}} = x^{\frac{2}{2}} = x$ because of the exponents multiplying together?  Also, doesn't $(\sqrt{x})^2$ preserve the sign of $x$? But shouldn't $(\sqrt{x})^2 = (\sqrt{x})(\sqrt{x}) = \sqrt{x^2}$?  How do I reconcile all this? What rules am I not aware of?  Edit: Since someone voted to close my question, I should probably explain the difference between my question and , regardless of how much I think the difference should be obvious to anyone who reads the questions. Cole Johnson was asking if there's any way to prove that $\sqrt{x^2} = |x|$. I am not asking that; I already accept the equation as fact. I'm asking how to resolve some apparent contradictions that arise when considering square roots of squares, and how I should approach these types of problems. (Cameron, please read.)
108089
Prove that if m is a square integer then m is neither congruent to 2 modulo 5 nor congruent to 3 modulo 5.  I've seen this problem done for modulo 4 and modulo 8 but this doesn't seem to work for me as the 2k doesn't square nicely to a 4. Maybe I'm misunderstanding, but I'm very lost on this problem and where to start/how to prove. Case proof? Proof by contradiction? Any help is appreciated, thanks
104050
My problem is "Prove $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}$ is a simple extension"  I met a similar problem "Prove $\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}$ is a simple extension" and I showed that it is equal to Prove $\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}$. I let $u=\sqrt{2}+\sqrt{3}$ and deduce that $u-\sqrt{2}=\sqrt{3})$, then square it and prove $\sqrt{2} \in Q(\sqrt{2}+\sqrt{3})$. However, the same technique cannot be applied in the first problem since we have 3 square root.  I actually found a proof for the first problem, by letting $u=\sqrt{2}+\sqrt{3}+\sqrt{5}$ and calculating $u,u^2,..., u^5$ and prove $\sqrt{2},\sqrt{3}\in Q(u)$. Though it's not so nice. Looking for a simpler answer
95318
I'm trying to show that if $\alpha$ is the square root of a non-perfect square $d$, then for any rational $p/q$ there's a positive real $x$ such that $|\alpha - p/q| &gt; x/q^2$. I'm having trouble starting on the problem, could someone give a hint? Thanks
77712
Show that  $\tan {\pi  \over 8} = \sqrt 2  - 1$, using the identity $\tan 2\theta  = {{2\tan \theta } \over {1 - {{\tan }^2}\theta }}$    Using $\tan 2\theta  = {{2\tan \theta } \over {1 - {{\tan }^2}\theta }}$ with $\theta  = {\pi  \over {16}}$:  $\eqalign{   &amp; \tan {\pi  \over 8} = {{2\tan {\pi  \over {16}}} \over {1 - {{\tan }^2}{\pi  \over {16}}}}  \cr    &amp; \tan {\pi  \over 8} = 0.41421.... \cr} $  I essentially get an answer equivalent to $\sqrt 2  - 1$ but not an "exact" answer, how do I go about this question to get a result in "exact" form?  Thank you.
139
How do I prove $\sqrt{x + y} \le \sqrt{x} + \sqrt{y}$? for $x, y$ positive?  This should be easy, but I'm not seeing how.  A hint would be appreciated.
69530
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.
100900
Prove $a^2$ is even $\implies$ $a$ is even.  It was proven via contradiction by my friend.  Here is the proof in question, which uses proof by contradiction  Assume towards a contradiction that $a$ is odd.  Let $a = 2k+1$, then $a^2 = 4k^2 + 4k + 1 = 2(2k^2+2k)+1$ Therefore $a^2$ is odd.  Contradiction.  To me, this proof in question seems incorrect because it shows that if $a$ is odd then $a^2$ is odd. We have shown that the converse of what we want to contradict is a contradiction.  I feel like this proof by contradiction is not mathematically correct or vigorous enough, and for it to be correct we would have to start from $a^2$ being even, working our way down to $a$ being odd, and then finding a contradiction.    Am I correct? Or am I just confused about something?
16666
Prove that for for all $p,q\in \mathbb{Z}$, $q&gt;0$ we have: $$ \left| \sqrt{2}-\frac{p}{q} \right| &gt; \frac{1}{3q^2}. $$  To be honest, I do not know where to start - any help would be appreciated.
70752
There is a problem I'm trying to solve from the Polish XI Olimpiada Matematyczna Gimnazjalistów. It is question 2 from . (The deadline of 12 October 2015 has passed.)    So far I found $x^2$ and $y^2$ (that are $|AP|^2$ and $|PD|^2$).  In order to prove, that AP=2PD, can I just divide $$ \sqrt{\frac{x^2}{y^2}} $$ and show, that it equals $2$?