Here is a conundrum I pondered upon while solving an equation -  2x2 , generally means we are adding 2 two times , that is , 2+2  Then ,  (-2)x(-2) , should go like this (-2)+(-2)=-4 ,  But , that's not true , the answer is +4 .  Can anybody figure out and help me with it..
123358
How to logically understand the multiplication of two integers ?  Eg:  3 x 4 = 12 (is understandable)  -3 x4 = -12 (is also somewhat understandable)  But , 3 x -4 = -12 (is NOT understandable)  -3 x -4 = 12 (is also NOT understandable)  Now , at this we must assume that commutative property may not be true .  What is the logical explanation for something being multiplied negative times ?
79941
With $i^2=-1$, what is $i!$ ?  I'm not sure if it is $i$ or if it is an infinite product $i(i-1)(i-2)(i-3)(i-4)(i-5)\dots$  It would make sense either way.  Is it not defined?
32657
What are the solutions in positive integers of the equation:  $${1+2^x+2^{2x+1}=y^2}$$ I tried to factorize the equation but it didn't help much. Clearly $y $ is an odd integer. Substituting $y =2n+1$, we get  $2^x+2^{2x+1}=(2n)\cdot{(2n+2)}$ $\Rightarrow (2^{x-2})\cdot(1+2^{x+1})=(n)\cdot(n+1)$ Which is the product of 2 consecutive integers. Does it help? I don't know.
20997
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
90423
Good-morning Math Exchange (and good evening to some!)  I have a very basic question that is confusing me.   At school I was told that $\sqrt {a^2} = \pm a$  However, does this mean that $\sqrt {a^2} = +2$ *and*$-2$ or does it mean: $\sqrt {a^2} = +2$ *OR*$-2$  Is it wrong to say 'and'? What are the implications of choosing 'and'/'or'  Any help would be greatly appreciated. Thanks in advance and enjoy the rest of your day :)
16867
I've tried two ways, but get stuck.  I've tried to simplify, but didn't know what to do next, and I've tried to solve it like a Quadratic equation but got stuck too.   One way got me this:  $$\frac{z}{2} \times (-1+3i+z)-1-i=0$$ - don't know what next.  option b, got me mess, while trying to make the equation into a Quadratic.
58522
Possible Duplicate:          I'm trying to understand the exact point of failure in the following reasoning:  \begin{equation*} 1 = \sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{\sqrt{-1}^2\sqrt{-1}^2} = \sqrt{(\sqrt{-1}\sqrt{-1})^2} = \sqrt{-1}\sqrt{-1} = \sqrt{-1}^2 = -1. \end{equation*}  I've been previously told that the problem is due to square root not being a function in C; which I found totally unhelpful. Could someone please explain the problem here in simpler terms.  Edit:  Thank you all for your comments in trying to help me understand this. I finally do. Following is the explanation on my problems in understanding this, in case it'll be of any help to anyone else.  My problem was really due to using an incorrect definition of i: $i = \sqrt{-1}$. While the correct definition would be: $i^2 = -1$.  My incorrect definition led me reasoning, such as (which superficially seemed to give expected results. I see now that this is incorrect, too):  \begin{equation*} \sqrt{-9} = \sqrt{9 * (-1)} = \sqrt{\sqrt{9}^2 \sqrt{-1}^2} = \sqrt{(\sqrt{9} \sqrt{-1})^2} = \sqrt{9} \sqrt{-1} = 3i. \end{equation*}  Instead, had I used the correct definition of i:  \begin{equation*} {(xi)}^2 = x^2i^2 = -x^2 = -9, \\ x^2 = 9, \\ x = +- 3. \end{equation*}  Now, analyzing the equations in the original problem, I can see at least the following two errors:  1) In the third =, I'm relying on $-1 = {\sqrt{-1}}^2$, while I should be relying on: $-1 = (+-\sqrt{-1})^2$ which would of course give two different branches. Hmm.. on the second reading, this isn't really a problem, as even with the two separate branches, both of them will lead to the result in the next step.  2) In the fifth =, I'm relying on $\sqrt{i^4} = i^2$, which would be correct, if i was a non-negative number in R. But as i is the imaginary unit and in C: $\sqrt{i^4} = \sqrt{i} = +-(1/\sqrt{2})(1 + i)$.
24417
I have the equation $x^2=x$.  If I divide $x$ from both sides I get $x=1$.  Yet clearly $x$ can also equal $0$.  What step in this process is wrong? It seems to me that there's only one step. And isn't dividing the same thing from both sides a valid step?  I hope this isn't a stupid question because I feel dumb asking about something so basic.  EDIT: To clarify, what I'm looking for is not only an explanation for why my methodology is wrong, but also a better methodology that will keep me from missing possible solutions in the future.
1237
Ok i feel like this is an easy question that I'm making more complicated than it needs to be  $$ \begin{align} 2n+1 &amp;= (2n-1) + 2\\ (2n-1) &amp;= 2(n-1)+1 \end{align} $$  so is it $1$ ?
65082
Question: simplify $$\frac{(n-1)!}{(n-2)!}$$   What I did was:  $$\frac{(n - 1)!}{(n - 2)! \times (n - 3)!}$$ This I did following the rule $n! = n \times (n - 1)!$.  But my answer just doesn't look correct and I don't have a solution guide that tells me the correct answer. This is why I don't know whether my answer is correct or not.
91397
Find all integer roots of: $x^2(y-1)+y^2(x-1)=1$   Obviously $(2,1)$ and $(1,2)$ are two answers. But I was unable to manipulate the equation algebraically giving a useful form for finding all other possible answers! I also tried to view it as a quadratic equation in terms of $x$, but forming the Delta of that equation didn't help much other than if $y=1$ then certainly Delta would be a square and equation would have answer for $x$...
84140
Is there a method to find which numbers to use when simplifying quadratics?   For example $x^2 + 5x + 6$ is easy enough to find, but what if I have bigger numbers, or I have this quadratic expression: $x^2 + 7x – 6$, then it gets very hard to think of a valid simplification.  EDIT: Here's an example: $(x^2 + 5x + 6)/(x + 2)$. When simplifying the first half we get $(x + 2)(x + 3)/(x + 2)$, then we can cancel out $(x + 2)$ and the answer would be $(x + 3)$. The question is: How to always find the right numbers which to put in? (in this example they where 2 and 3)
100892
I have the equation: $$y^2=x^3-2$$ It seems to be deceivingly simple, yet I simply cannot crack it. It is obviously equivalent to finding a perfect cube that is two more than a perfect square, and a brute force check shows no solutions other than $y=5$ and $x=3$ under 10,000.  However, I can't prove it.   Are there other integer solutions to this equation? If so, how many? If not, can you prove that there aren't?   Bonus: What about the more general equation" $$y^2=x^3-c$$ Where $c$ is a positive integer?
46796
I'm struggling to solve this problem:  "If the equation $ax^2 + bx +c$ $(a \neq 0)$ admits real and not null roots of $x_1$ and $x_2$, obtain the equation which evaluates to roots $(x_1)^2$ and $(x_2)^2$"  I know the answer is $a^2x^2 -(b^2 - 2ac)x + c^2$ but I am not able to get to this answer, I was trying to derive it from the sum and product equations $S=-b/a$ and $P=c/a$ but the answer I'm getting to is a monstruous different thing.  Could someone explain to me how to get to this answer?  Thanks
92099
Find all real solutions for $x$ in $2(2^x−1)x^2+(2^{x^2}−2)x=2^{x+1}−2$.   I started off by dividing $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2$ by $2$, and I got $(2^x-1)x^2 + (2^{x^2-1}-1)x = 2^x -1$. I tried dividing by $2^x-1$ on both sides which would give me a simple quadratic to solve for $x$, but I don't know how to simplify $\frac{(2^{x^2}-1)}{(2^x-1)}$. Am I missing a simple trick here, or am I on a completely wrong path to solving this?
59648
im not good at formatting. May I know why it’s false   $x^2= x+x+x+\ldots(x\; \text{times})$  apply derivative on both sides  $=&gt; \frac{d}{dx}(x^2)=\frac{d}{dx}(x+x+x+\ldots(x\; \text{times}))$  $=&gt; 2x=1+1+1+\ldots(x\; \text{times})$  $=&gt; 2x=x$  $=&gt; 2=1$
101477
Why does the equation $x^2\equiv2 \pmod 5$ have no solutions?  I did a remainders table and found that $$x^2\equiv0;1;4\pmod 5$$  But is there any way to justify this besides that?  The original equation was $2x^2\equiv9\pmod 5$ but I got it to the form above.