I have seen this image and surprised that we can prove $2 + 2 = 5$. can any one tell me what is wrong with this image.     Prove that, $2+2=5$.      We know that, $2+2=4$      $$\begin{align}\Rightarrow2+2&amp;=4-\dfrac92+\dfrac92\\\,\\ &amp;=\sqrt{\left(4-\dfrac92\right)^2}+\dfrac92\\ &amp;=\sqrt{16-2\cdot4\cdot\dfrac92+\left(\dfrac92\right)^2}+\dfrac92\\ &amp;=\sqrt{-20+\left(\dfrac92\right)^2}+\dfrac92\\ &amp;=\sqrt{(5)^2-2\cdot5\cdot\dfrac92+\left(\dfrac92\right)^2}+\dfrac92\qquad\qquad\qquad\\ &amp;=\sqrt{\left(5-\dfrac92\right)^2}+\dfrac92\\ &amp;=5-\dfrac92+\dfrac92\\ &amp;=5\\\,\\&amp;\therefore\,2+2=5\text{ (Proved)}\\ \end{align}$$
20466
$$\begin{align} 2+2 &amp;= 4 - \frac92 +\frac92\\ &amp;= \sqrt{\left(4-\frac92\right)^2} +\frac92\\ &amp;= \sqrt{16 -2\times4\times\frac92 +\left(\frac92\right)^2} + \frac92\\ &amp;= \sqrt{16 -36 + \left(\frac92\right)^2} +\frac92\\ &amp;= \sqrt {-20 +\left(\frac92\right)^2} + \frac92\\ &amp;= \sqrt{25-45 +\left(\frac92\right)^2} +\frac92\\ &amp;= \sqrt {5^2 -2\times5\times\frac92 + \left(\frac92\right) ^2} + \frac92\\ &amp;= \sqrt {\left(5-\frac92\right)^2} +\frac92\\ &amp;= 5 + \frac92 - \frac92 \\ &amp;= 5\end{align}$$  Where did I go wrong
94373
Prove that $n$ is even   $$\left(\dfrac {\cos A + \cos B}{\sin A-\sin B}\right)^n + \left(\dfrac {\sin A+\sin B}{\cos A- \cos B}\right)^n = 2\cot^n \left(\dfrac {A-B}{2}\right)$$  Please help. I didn't get any idea regarding this.
90592
how would you prove that $2\sqrt5 +\sqrt{11}$ is irrational? I started with a proof by contradiction that assumes that $2\sqrt5 +\sqrt{11}$ is rational and therefore there exist integers $a$ and $b$ such that $\frac{a}{b}=2\sqrt5 +\sqrt{11}$ and squaring both sides yields $\frac{a^2}{b^2}=31 +4\sqrt5\sqrt{11}$ and from this point im stuck as i dont know how to continue to arrive at a contradiction.
38532
I'm trying to understand the following proof of why      No two Fermat numbers have a common divisor greater than $1$.   Hardy and Wright show the following proofs:   If $x = 2^{2^n}$, we have  $$\frac{F_{n+k} - 2}{F_n}(\circ) = \frac{2^{2^{n+k}}-1}{2^{2^n}+1} = \frac{x^{2^k}-1}{x+1} = x^{2^k-1} - x^{2^k-2} + \ldots -1$$  The main issue I'm having is the $\circ$ part. I don't understand how they get that. The rest of the proof makes sense but the first part is really confusing me. Can someone please explain where they get that fraction?  Thanks a bunch!
114930
This is a question the prof asked in my university math proofs course. I've tried this for a bit over an hour and I am stuck.   The method I personally chose was proof by cases:  Case 1: n is even:    Thus $n = 2k$ for some k ∈ Z. So $n = 2k = 5a + 2b$  Here I can choose $a=0$, thus $n = 2k = 2b$ Therefore, if n is even there always exists some a,b∈ Z such that $n = 5a + 2b$ and case 1 is true.  Case 2: n is odd:  Thus $n = 2k+1$ for some k ∈ Z. So $n = 2k+1 = 5a + 2b$  This is where I'm stuck, I cant seem to show that the case is always true for when n is odd.  If you could give me a hint in the right direction or complete my method of proof by cases that would be great, or if you had another method of proving it that would be helpful too.  Thanks, John
17489
Prove: If $a$ and $b$ are odd integers, then $8\mid (a^2-b^2)$  So far I have: $$8\mid \left((2n+1)^2-(2m+1)^2\right)\Longleftrightarrow 8\mid \left(4n^2+4n+1-4m^2-4m-1\right) $$ Is this right so far? Where do I go from here? Step by step explanation please!
70109
$$ z = 1+2i \ (complex \ number) \\ z^n = a_n + b_ni \ \ \ (a_n, b_n \in \mathbb{Z}, n \in \mathbb{N}^*) $$ Prove that $ b_{n+2} - 2b_{n+1} + 5b_n = 0$  How can I solve this? Thank you!  EDIT: Or please tell me your ideas.
74361
Ok, so I have the following proof. Let $z$ and $w$ be complex numbers. Prove $\lvert z+w \rvert ^2 + \lvert z-w \rvert^2 = 2[\lvert z \rvert^2 + \vert w \rvert^2]$. Using $\vert z \rvert^2=z\bar{z}$, I have said   $\lvert z+w \rvert ^2 + \lvert z-w \rvert^2 = 2(\lvert z \rvert^2) + \vert w \rvert^2)$  $(z+w)\bar{(z+w)}+(z-w)(\bar{(z-w)}$  $(z+w)(z-w)+(z-w)(z+w)$  $z^2 - w^2 + z^2 - w^2$  $z^2 + z^2 + -w^2 + -w^2$  $2(z^2) + 2(-w^2)$  $2[z^2 + (-w)^2]$  $2[\lvert z \rvert^2 + \vert w \rvert^2]$  I guess my question is, is my logic correct? I am not so skilled with proof writing and though I feel that this is correct, I have been known to be wrong in the past. Just need some insight. Thank you.
110441
I know how to prove this by induction but the text I'm following shows another way to prove it and I guess this way is used again in the future. I'm confused by it.  So the expression for first n numbers is: $$\frac{n(n+1)}{2}$$  And this second proof starts out like this. It says since:  $$(n+1)^2-n^2=2n+1$$  Absolutely no idea where this expression came from, doesn't explain where it came from either.  Then it proceeds to say: \begin{align} 2^2-1^2&amp;=2*1+1 \\ 3^2-2^2&amp;=2*2+1\\ &amp;\dots\\ n^2-(n-1)^2&amp;=2(n-1)+1\\ (n+1)^2-n^2&amp;=2n+1 \end{align} At this point I'm completely lost. But it continues to say "adding and noting the cancellations on the left, we get" \begin{align} (n+1)^2-1&amp;=2(1+2+...+n)+n \\ n^2+n&amp;=2(1+2+...+n) \\ (n(n+1))/2&amp;=1+2+...+n \end{align}  Which proves it but I have no clue what has happened. I am entirely new to these math proofs. Im completely lost. I was great at high school math and calculus but now I haven't got the slightest clue of what's going on. Thanks
68048
We have to prove that  "the product of two numbers ending in 6 also ends with 6" mathematically.  I have no clue how to start. I don't want you to prove it for me! but some hints would be very helpful as I'm totally stuck on how to start this prove.
63744
Prove that:  $$a^{2^n} \equiv 1\pmod{2^{n+2}}$$  I have done the base case, and then have an assumption that:  $$a^{2^k} \equiv 1 \pmod {2^{k+2}}$$  I wish to prove the last step, that:  $$a^{2^{k+1}} \equiv 1 \pmod{2^{k+3}}$$  How would I do this?
87766
Let $a$, $b$, and $c$ be integers, where a $\ne$ 0. Then $$ $$ (i) if $a$ | $b$ and $a$ | $c$, then $a$ | ($b+c$) $$ $$ (ii) if $a$ | $b$ then $a$|$bc$ for all integers $c$; $$ $$ (iii) if $a$ |$b$ and $b$|$c$, then $a$|$c$.  Prove that if $a$|$b$ and $b$|$c$ then $a$|$c$ using a column proof that has steps in the first column and the reason for the step in the second column.  My book is really vague. I'm not really sure what to do..  First I was thinking something like this: $a\mid b\Rightarrow b=as$ and  $b\mid c\Rightarrow c=bt$ and $a\mid c\Rightarrow c = au$ $$ $$ $b + c + c = b + 2c$  $$ $$ $=as + 2(bt+au)$  Yea then I just got lost.
105982
Find all $x \in (-1, +\infty)$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$.  What I have done so far was a substitution $y = x + 2$ which results in a nicer form of the equation:   $(y - 1)^{y - 2}(y + 1)^{y - 2} + 2^{y - 2}y^{y - 2} = (y^2 + 1)^{y - 2}$ and $y &gt; 1$, which can then be rewritten as:  $$ \left( \frac{y^2 + 1}{2y} \right)^{y - 2} - \left( \frac{y^2 - 1}{2y} \right)^{y - 2} = 1 $$ Also, my intuition is that $y = 4$ is the unique solution and initially I wanted to prove that this function of $y$ is injective, but it didn't work either.  Do you have any suggestion on how I can continue?
82780
To prove (1)  $$\sum_{i=1}^n \dfrac {i}{2^i} &lt;2$$  Someone shows that (2)  $$\sum_{i=1}^n \dfrac {i}{2^i} =  2-\dfrac{n+2}{2^n}$$  so we can prove (1).  Can anyone guide me how from (1) they come up with (2) please ? On what base they find (2) ?
115709
Proving by induction. We'd like to show that $2 + 4 + 6 + \cdots+ 2n = n(n + 1)$.  A nice way to do this is by induction. Let $S(n)$ be the statement above. An inductive proof would have the following steps:  Show that $S(1)$ is true.  Show that if $S(1),\ldots,S(k)$ are true, then so is $S(k + 1)$.    This question is really starting to bug me, am I doing something wrong or is the equation wrong for that series? It only seems to work for $S(1)$ but after that it does not give the correct series. The $n(n+1)$ series is $2, 6, 12,\ldots.$ I'm really confused can someone please nudge me in the right direction (I know how proofs by induction works) I'm just having problems with this one in particular.
99875
Let $a,b,c,d\in \mathbb{R}$ and $a+b+c+d=2$. Prove that $$\frac{a}{a^2-a+1}+\frac{b}{b^2-b+1}+\frac{c}{c^2-c+1}+\frac{d}{d^2-d+1}\le \frac{8}{3}.$$   We have $$\frac{a}{a^2-a+1}\le \frac{4}{3}a\Longleftrightarrow \frac{-a\left(2a-1\right)^2}{3\left(a^2-a+1\right)}\le 0,$$ (Right)  So $$LHS\le \frac{4}{3}\left(a+b+c+d\right)=\frac{8}{3}=RHS$$  When $a=b=c=d=\frac{1}{2}$  But also $a=b=c=1;d=-1$ i don't know how to solve it.
2019
I have been boggled by this question for a while as well. Prove that $$(2a+b)(2b+a)=2^c$$  Is impossible.  I know that if a and b do exist then they must be even. I am trying to use this fact to contradict the statement. I haven also tried rewriting a and b as products of powers of twos and a odd factor