Basically, there is no error in the following steps(as it seems), but there is some error due to which 2 + 2 = 5. What is it?                        -20 = -20                     16-36 = 25-45              16-36+(81/4) = 25-45+(81/4) (4^2)-(2*4*9/2)+((9/2)^2) = (5^2)-(2*5*(9/2))+((9/2)^2)               (4-(9/2))^2 = (5-(9/2))^2                   4-(9/2) = 5-(9/2)                         4 = 5                       2+2 = 5
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
87460
I want to show rigorously that this is 2. I'm sure there's a faster way than by trying to see if $\sqrt{5} = a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6}$ (linear combination of the basis elements for $\mathbb{Q}(\sqrt{3}, \sqrt{2})$). I'm just not sure what.
47399
I found crazy (for me at least) induction example, in fact it just would be nice to prove. (Even have problems with starting) Any hints are highly valued: $$0^2\binom{n}{0}+1^2\binom{n}{1}+2^2\binom{n}{2}+\cdots+n^2\binom{n}{n}=n(1+n)2^{n-2} $$
47495
We know that 2 x 2 /2 can be solved by removing '2' from the denominator and numerator, we can't do the same if the operation was addition, These "rules" have been established based on the understanding of these operations and how these interact with each other.  Up until now, I've just been accepting it as a rule, but this causes a lot of discomforts while doing math.  I generally get good grades in math, but that's only by following these rules and patterns but there is no genuine understanding as to why we can remove "2" from the numerator and denominator in 2 x 2 /2 but we cannot in 2 + 2/2.  Secondly, 2 x 2 / 2 can be solved by removing "2" from the numerator and the denominator and it gives the same result as 4/2, how?. Is it an axiom?  I hope to gain an intuition for the above concepts, so I can be free from my long-held guilt.  Thanks.
54447
If you know it, also try to include the precise reason why the proof is fallacious. To start this off, let me post the one that most people know already:    Let $a = b$.  Then $a^2 = ab$  $a^2 - b^2 = ab - b^2$  Factor to $(a-b)(a+b) = b(a-b)$  Then divide out $(a-b)$ to get $a+b = b$  Since $a = b$, then $b+b = b$  Therefore $2b = b$  Reduce to $2 = 1$   As @jan-gorzny pointed out, in this case, line 5 is wrong since $a = b$ implies $a-b = 0$, and so you can't divide out $(a-b)$.
5958
I was reading Ian Stewart's Concepts of Modern Mathematics.     Using congruences, It's possible to explain why all perfect squares end in $0,1,4,5,6,9$ but not in $2,3,7,8$.    With this I had the idea of exploring the congruences for both sides of $n^2=2m^2$ in Mathematica:  Table[Mod[n^2, 9], {n, 0, 20}]  Table[Mod[2 m^2, 9], {n, 0, 20}]  And had the results:  {0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 1, 4, 0, 7, 7, 0, 4, 1, 0, 1, 4}  {0, 2, 8, 0, 5, 5, 0, 8, 2, 0, 2, 8, 0, 5, 5, 0, 8, 2, 0, 2, 8}  But I'm still not sure if the outputs really show what I'm looking for, I have also tried $mod \;10$. The idea is still pretty loose in my mind, I'm stuck on deciding if this proves something or what directions I could take in this enterprise.
89373
I cannot complete the third step of induction for this one. The assumption is $3^{2n+3}+40n-27=64k$, and when substituting for $n+1$ I obtain $3^{2n+5}+40n+13=64k$. I've tried factoring the expression, dividing, etc. but I cannot advance from here.
26390
Is there an error in this question?  Let u = $(1,0,2)$, v = $(2,1,0)$ and w = $(0,2,1)$. Compute  ||u + v - w.||v = v  This is exactly what is in the question (even the dot after w which I don't know what it means).  Thanks
4300
I know there must be something unmathematical in the following but I don't know where it is: \begin{align} \sqrt{-1} &amp;= i \\\\\ \frac1{\sqrt{-1}} &amp;= \frac1i \\\\  \frac{\sqrt1}{\sqrt{-1}} &amp;= \frac1i \\\\  \sqrt{\frac1{-1}} &amp;= \frac1i \\\\  \sqrt{\frac{-1}1} &amp;= \frac1i \\\\  \sqrt{-1} &amp;= \frac1i \\\\  i &amp;= \frac1i \\\\  i^2 &amp;= 1 \\\\  -1 &amp;= 1 \quad !!? \end{align}
92526
How can I get the numbers from (0-30) by using the number $2$ four times.Use any common mathematical function and the (+,-,*,/,^) I tried to solve this puzzle, but I couldn't solve it completely. Some of my results were: $$2/2-2/2=0$$ $$(2*2)/(2*2)=1$$ $$2/2+2/2=2$$ $$2^2-2/2=3$$ $$\frac{2*2}{2}+2=4$$ $$2^2+2/2=5$$  $$2^2*2-2=6$$ $$\frac{2^{2*2}}{2}=8$$ $$(2+2/2)^2=9$$ $$2*2*2+2=10$$ $$2*2*2*2=16$$ $$22+2/2=23$$ $$(2+2)!+2/2=25$$ $$(2+2)!+2+2=28$$
21576
Given $5x/30x^2$ I was wondering which is the correct equivalent form.  According to BEDMAS this expression is equivalent to $5*\cfrac{x}{30}*x^2$ but, intuitively, I believe that it could also look like: $\cfrac{5x}{30x^2}$  I asked this question on MathOverflow (which was "Off-topic" and closed) and was told it was ambiguous. I was wondering what the convention was or if such a convention exists. According to Wikipedia the .
76244
Let $S(n)$ be the proposition "$2^n &gt; n^2$"  Basis step: $S(5)$ is true, since $2^5 = 32$, $ 32 &gt; 25 = 5^ 2$  Inductive step:  Induction hypothesis:   Assume $S(K)$ is true, $2^K &gt; K^2$.  Then  $2^{K+1} = 2.2^K &gt; K^2 + K^2$   I don't know how to proceed after this. How to complete this proof?
123693
Yes, there is no one who doesn't know this problem.My question is only about curiosity.     $$C(n) = \begin{cases} n/2 &amp;\text{if } n \equiv 0 \pmod{2}\\ 3n+1 &amp; \text{if } n\equiv 1 \pmod{2} .\end{cases}$$   On this problem, I caught something like this.I'm sure, We all realized that.  For example, $n=19$, we have $6$ odd steps.  We know that, even steps are not important, because each even number is converted to an odd number.     $19\Longrightarrow 29 \Longrightarrow 11\Longrightarrow 17 \Longrightarrow13 \Longrightarrow 5 \Longrightarrow 1$   Then, for $n=77$, We have also $6$ odd steps.     $77\Longrightarrow 29 \Longrightarrow 11\Longrightarrow 17 \Longrightarrow13 \Longrightarrow 5 \Longrightarrow 1$   For $n=9$      $9\Longrightarrow 7 \Longrightarrow   11 \Longrightarrow 17 \Longrightarrow 13\Longrightarrow 5 \Longrightarrow 1$   Again we have $k=6$ odd steps.     I want to know / learn / ask, for $k=6$, (Generalized: for any number $k$ ) can we produce a formula(s) to catch all such numbers, which gives the result  $1$?    Thank you!
92761
Could someone please explain how to solve this : $x^4 - 10x^3 + 21x^2 + 40x - 100 = 0$ - not the answer only, but a step-by-step solution.  I tried to solve it, with the help of khanacademy, but still I have no idea how to correctly solve it.  Thank you so much in advance!
46053
I'm not sure if it's correct, but what I have so far is; $$21n^5 + 10n^3 + 14n ≡ (1 + 0 - 1) ≡ 0 \mod 5$$ but I'm having trouble solving it in $\bmod 3$. I have: $$21n^5 + 10n^3 + 14n ≡ (0 + (?) + 2),$$ I'm not sure how to solve the $10n^3$ part of the congruence in $\mod 3$. Also, if I have any mistakes please let me know.
51415
I reduced the equation:  $84x \equiv 68\mod 400$  $21x \equiv 17\mod 100$  $100 = 4 \times 21 + 16$  $21 = 1 \times 16 + 5$  $16 = 3 \times 5 + 1$  so we have...  $16 = 100 - 4 \times 21$  $5 = 21 - 1 \times 16$  $1 = 16 - 3 \times 5$  Now, I need the equation in the form of $21v + 100w = 1$ and I have to find v and w  $1=16-3 \times 5$  $=16-3(21-1 \times 16)$  $=16 - 3 \times 21 +3 \times 16$  $=100 - 4 \times 21 -3 \times 21 +3 \times 16$  At this point I don't know what to do. Please help.
60543
I'm not entirely sure if I'm going about proving $n^2+n$ is even for all the natural numbers correctly.  $P(n): = n^2+n$     $P(1) = 1^2+1 = 2 = 0$ (mod $2$), true for $P(1)$   Inductive step for $P(n+1)$:     $\begin{align}P(n+1) &amp;=&amp; (n+1)^2+(n+1)\\ &amp;=&amp;n^2+2n+1+n+1\\ &amp;=&amp;n^2+n+2(n+1)\end{align}$   Does this prove $n^2+n$ is even as it's divisible by $2$? Thanks!