$$ a = b $$ $$ a^2 = ab $$ $$ a^2 - b^2 = ab - b^2 $$ $$ (a+b)(a - b) = b(a - b) $$ $$ a + b = b $$ $$ 2b = b $$ $$ 2 = 1 $$  Does the self-reference to the original formula make this path of argument invalid? I am confused as to what's going on here.  Would this be a better continuation from $a + b = b$:  $$ a + b = b \implies a = 0 \implies b = 0$$
30704
I have a "proof" that has an error in it and my goal is to figure out what this error is. The proof:  If $x = y$, then  $$ \begin{eqnarray} x^2 &amp;=&amp; xy \nonumber \\ x^2 - y^2 &amp;=&amp; xy - y^2 \nonumber \\ (x + y)(x - y) &amp;=&amp; y(x-y) \nonumber \\ x + y &amp;=&amp; y \nonumber \\ 2y &amp;=&amp; y \nonumber \\ 2 &amp;=&amp; 1 \end{eqnarray} $$    My best guess is that the error starts with the line $2y = y$. If we accept that $x + y = y$ is true, then   $$ \begin{eqnarray} x + y &amp;=&amp; y \\ x &amp;=&amp; y - y \\ x &amp;=&amp; y = 0 \end{eqnarray} $$  Did I find the error? If not, am I close?
12345
Matrix Question  If $A^2 = I$, $B^2=I$ and $(AB)^2=I$, then $AB = BA$  Basically, got up to $A(BA-AB)B = 0$ (by cancelling and equating terms from $I^2 = I$ and to $A^2B^2 = A^2B^2$ and using distributive laws), but that doesn't work out too well!  Thanks for help in advance!
72859
Prove that $| |a|-|b| | \le |a-b|$  (1) I square both sides since I know they are positive:  $$a^2-2|ab|+b^2 \le a^2-2ab+b^2$$ $$|ab| \ge ab$$ Let $m=ab$ $$|m| \ge m$$ Which is part of the definition of the absolute value.  Is it enough or I need to rewrite my proof?
69330
The natural numbers $a$ and $b$ are such $a^2+ab+1$ is divisible by $b^2+ba+1$. Prove that $a = b$.   I tried to algebraically manipulate it as follows:  $(b^2 + ba + 1)k = a^2 + ab + 1$  $[b(a + b) + 1]k = a(a + b) + 1$  $kb(a + b) + k = a(a + b) + 1$  $k - 1 = (a - kb)(a + b)$  I'm stuck here. What should I do next? A case-by-case analysis of possible congruencies would be too tedious and inelegant.
38871
I'm having trouble deriving the following fact from the basic properties of $+_{\mathbb{N}}$ and the definition:  Definition. $\forall n, m \in \mathbb{N}: (n \geq m) \leftrightarrow (\exists a \in \mathbb{N}): n = m + a$. Fact. $\forall a, b \in \mathbb{N}: (a \geq b ~\land~ b \geq a) \rightarrow a = b$.  I've tried several different strategies, but none works. I think the fact that 0 isn't the successor of any natural number is the key both in the base case and the inductive step, but I keep going in circles trying to make use of that (and other basic facts about addition on $\mathbb{N}$). I would show you my notes, but there is really nothing of substance in them. I've made almost no progress since yesterday.
102197
Theorem. $|a||b|=|ab|$  Proof. Applying the definition of absolute value, the left hand side of the equation could be either $a\times(-b)$ or $(-a)\times(b)$ or $a\times b$ or $(-a)\times(-b)$. For this reason we could have either $-ab$ or $ab$.  If we apply the definition of absolute value to the right hand side of our equation, we discover that we could have either $-ab$ or $ab$. So the theorem is proved.  Is this proof valid? Is it elegant from a mathematical viewpoint?  Thank you.  Edit: The definition of absolute value I am using above is the following: $$|x|=\begin{cases}x, &amp; \text{if $x\ge0$}\\-x,&amp;\text{if $x&lt;0$}\end{cases}$$
65318
$T:\mathbb P_2[x]↦\mathbb P_2[x]$ s.t.:   $T(1)=3+2x+4x^2$, $T(x)=2+2x^2$, $T(x^2)=4+2x+3x^2$.   Is there base $B$ of $\mathbb P_2[x]$ that $[T]_B$= \begin{bmatrix}     2 &amp; 0 &amp; 0\\     -1 &amp; 1 &amp; 0\\     5 &amp; 0 &amp; 4\\  \end{bmatrix}  Please help. Thanks.
45673
What's wrong with this demonstration?:  $$A \iff 1 = 1^1$$ $$A \implies 1 = 1^\frac{2}{2}$$ $$A \implies 1 = (1^2)^\frac{1}{2}$$ $$A \implies 1 = ((-1)^2)^\frac{1}{2}$$ $$A \implies 1 = (-1)^\frac{2}{2}$$ $$A \implies 1 = (-1)^1 = -1$$  So finaly we get  $$1 = -1$$
87756
This actually triggered me in my mind from . After some playing around I notice that the relation $a^{\log_b x} = x^{\log_b a}$ is true for any valid value of $a,b$ and $x$. I am very inquisitive to see how this holds ?
89874
$$\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$$  I have attempted this question by expanding the left side using the cosine sum and difference formulas and then multiplying, and then simplifying till I replicated the identity on the right. I am not stuck. What's bothering me is that the way I went about this question seemed like a rather "clunky" method. I'm just curious if I've missed some underlying pattern that could have made it easier to reproduce the identity on the right.  The way I did it:  $$\begin{array}{l} \cos (A + B)\cos (A - B)\\ \equiv (\cos A\cos B - \sin A\sin B)(\cos A\cos B + \sin A\sin B)\\ \equiv {\cos ^2}A{\cos ^2}B - {\sin ^2}A{\sin ^2}B\\ \equiv {\cos ^2}A(1 - {\sin ^2}B) - (1 - {\cos ^2}A){\sin ^2}B\\ \equiv {\cos ^2}A - {\cos ^2}A{\sin ^2}B - {\sin ^2}B + {\cos ^2}A{\sin ^2}B\\ \equiv {\cos ^2}A - {\sin ^2}B\end{array}$$
77653
For me, it is pretty clear that $a + \overline{a}*b = a + b$, because the first $a$ in the or will make sure that if the second term must be 'evaluated', $a$ will always be false, and therefore won't matter in the and - probably because I'm a programmer and it is very common to see this unnecessary cluttering in long if-else structures.  I can easily prove it with the truth table as well, just make both expressions', and turns out they are identical.  But I was looking for a 'algebraic' way to prove it. That is, something using the basic operations properties like the fact that they are transitive, commutative, associative, etc. For example, if I was to prove this other simple identity $a + a*b = a$ I would simply do something like:  $a + a*b = a*(1 + b) = a*1 = a$.  Is there a way to also prove the former using this algebraic operations, without resorting to the truth table?
98084
How can I prove the Boolean algebraic rule $$A+A'B=A+B$$ without using a truth table?  With the truth table, it is easy to see that the two are equal, but how can I prove it using lesser Boolean identities?
106525
I have this proposition to prove: The equation $x^2 = -1$ has no solution in $\mathbb Z$. I was told that this is an opportunity for a proof by contradiction. I have already proven that for $m \in\mathbb Z$, if $m \ne 0$ then $m^2 \in\mathbb N$. I have also proven that 1 is a natural number (hence, -1 isn't one).  Here is my strategy: assume that there is a solution. Hence $x^2$ has 3 options: \begin{align*} x^2 \in\mathbb N\\ -(x^2) \in\mathbb N\\ x^2 = 0 \end{align*}  Option 1 is not possible because -1 is not a natural number. Option 2 doesn't make sense because         -$\mathbb N \in \mathbb N$ and option 3 doesn't make sense either because $0 \ne -1$. What do you think? Thank you!    Here is a simpler strategy based on the comments below.  Proof: Assume that $x^2 = -1$ has a solution in $\mathbb Z$, then $x^2$ will either be $0$ or $\in\mathbb N$. However, $-1 \notin\mathbb N$ and $-1 \ne 0$. Hence, there is a contradiction. Here is how I have proven that $-1 \notin\mathbb N$: Let $m \in\mathbb N$. Hence:  \begin{align*} -m \notin\mathbb N\\ (-1)m \notin\mathbb N \end{align*}  If $-1 \in\mathbb N$, the product of $-1$ and $m$ should $\in\mathbb N$, which is not the case here. Hence, $-1 \notin\mathbb N$.
78026
In my math text book the proof for the Sum to Product rule starts with a substitution:  let A = (u+v)/2, let B = (u-v)/2  ...and then uses this substitution to prove that A+B = u  and  A-B = v  ...then substitutes (u+v)/2 and (u-v)/2 in the Product-to-Sum equation to demonstrate that the Sum-to-Product rule is true.  As it appears to me from what is presented in the text book, the substitutions  (u+v)/2 and (u-v)/2 were pulled out of thin air.  How was this determined to work? I doubt it was just a lot of guess and check until something worked.   Is there a point in math where one just needs to accept an equation for what it is and not worry about why or how it is?  I'm getting the feeling that understanding some of the how's and why's right now requires an understanding of math concepts that I haven't learned yet.  Any insight would be appreciated. Thanks!
81826
$(x+2)(x-1$)  Totally forgot how to do it. Isn't this the rule:  $(a+b)(c+d)$  $a*c +a*d +b*c +b*d$  The answer I'm getting is $x^2-x+2x-2$ which is wrong. Please help me out, thanks!
61648
Given matrices $A$ and $B$, where that $AB = A + B$, prove $AB = BA$. I keep coming up with AB = AB. It seems like basic algebra, but for the life of me, I'm getting nowhere :/. Someone help please?
57250
What I have done is this:  $a\equiv b \pmod{2n}$,  $a=b+c\times2n$, for some $c$,  $a^2=b^2+2b\times c\times2n+c^2\times2^2n^2$,  $a^2-b^2=(b\times c+c^2n)\times4n$, then  $a^2\equiv b^2\pmod{2^2n}$.  I think that this is right: what I DON’T understand is how to generalize this to:  $a\equiv b\pmod{kn}\Rightarrow a^k\equiv b^k \pmod{k^2n}$.  Please give me a hint.