Bogus Claim: If $a$ and $b$ are two equal real numbers, then $a = 0$  $a = b$  $a^2 = ab$  $a^2 - b^2 = ab - b^2$  $(a-b)(a+b) = (a-b)b$  $a + b = b$  $a = 0$  I found this in my proof handouts, and correct me if I'm wrong,but is it wrong because after line 4 we divide both sides by $(a - b)$ which would be $0$ if $a = b$ ?
121435
$$ a = b $$ $$ a^2 = ab $$ $$ a^2 - b^2 = ab - b^2 $$ $$ (a+b)(a - b) = b(a - b) $$ $$ a + b = b $$ $$ 2b = b $$ $$ 2 = 1 $$  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 $a + b = b$:  $$ a + b = b \implies a = 0 \implies b = 0$$
102543
I have a problem with this. Actually, still don't have the right way to start :/  Problem :  Let $A$ and $B$ be $n \times n$ complex matrices such that $AB - B^2A^2 = I_n$. Prove that if $A^3 + B^3 = 0$, then $BA - A^2B^2 = I_n$.  Thanks for any help.
72502
Show that for any real constants a and b, where b > 0,  $(n + a)^b = Θ(n^b)$.  So taking a stab at this.. I let $a = 0$, and $b = 1$  So I got $f(n) = (n + a)^b \implies f(n) = (n+0)^1$ which we know $\implies n^1$ or $O(n) \implies \Theta(n^1)$ ? I just took a guess. I'm not sure if that's even close to being right.
105031
I just started real analysis.  I don't have a background in proofs or logic, simply calculus.  So I'm trying to learn more about proofs--so forgive the basic question, please.  How do you go about proving this theorem: If $a$ and $b$ are any numbers, then there is one and only one number $x$ such that $a + x=b$.  This number is given by $x=b+(-a)$.  First part of the proof I understand--it's simple.  We simply use some axioms and do the following:  $a+b-a=b$, which of course is true.  This is just true from plugging in the $b+(-a)$ for $x$.  But how about the uniqueness issue?  In my text, it says:  $$(a+x)+(-a)=b+(-a) x=b+(-a)$$  How does this prove uniqueness?  Is this equation essentially saying that the ONLY possible value of $x$ is expressed by $b+(-a)$?  How would I have known to set this up as part of the proof if I was asked to prove something is unique?  Sorry for the basic question--have to start somewhere.
41613
Let  $$A = \{x | x = 2n+1, n\in\mathbb{Z}\}$$ and  $$B = \{x | x = 2m-21, m\in\mathbb{Z}\}.$$  I am trying to prove $A =B.$ I understand that I need to prove  $A\subseteq B$ and $B\subseteq A$; But my difficulty is how to prove it.  Any help would be appreciated!
43358
Let a and b be real numbers such that 0 &lt; a &lt; b. Prove $\frac{a+b}2 &gt; \sqrt{ab} &gt; \frac{2ab}{a + b}$  How can I prove this? Been working for hours and got nowhere. I see $\frac{a+b}{2}$ and $\frac{2ab}{a + b}$ are almost reciprocals. Is this important?  Please explain step by step.
2419
Prove $||a|-|b|| \leq |a-b|$  I do it by making 4 cases :  Case 1 : $ a \geq 0$, $ b\geq 0$  So we have   $||a||-||b||  = |a - b|$ . hence proved  Case 2 :   $a &lt; 0 , b&lt; 0$  $|-a+b| = |-(a-b)|=|a-b|$  Case 3  $a &gt;0, b&lt;0$  $||a||-||b|| =|a+b| \leq |a|+|b| \leq a -b \leq |a-b| $  Case 4:  $a &lt;0 ,b&gt;0$  $|-a+b| = |-(a-b)| = |a-b|$  Is this OK ? Thanks
54552
Prove or disprove: If a, b belong to the set of positive integers, and if a divides b and b divides a, then a=b. Does this hold if if a,b are not necessarily positive? Why or Why not?   Here is what I have:  If a and b are integers, we say a divides b if there exists an integer k so that b=ak. Thus:  $$b=ak$$  $$a=bj$$   $$a=akj$$ $$kj=1$$   Since a divides b and b divides a, we know that k,j must both be integer, which means k and j must both equal 1 or -1. This means:   $$a=b$$  This proves that a=b when we assume that a,b are both positive integers.   For the second part of the question, I am asked if I can still say a=b if the assumption about a, b is relaxed so that a,b belong to the integers (not the positive integers.) I think The answer is no, I cannot say that a=b for every case when we allow a,b to also be negative integers.   Here is my reasoning (I use the same idea that a=bj and b=ak here)   $$+/-a=akj$$  $$+/-1=kj$$   So $ |a|=|b|$ but not $a=b$  I'm very new to proofs so any suggestions or thoughts are greatly appreciated.
74884
Why is it that $$ab=ba$$ for positive integers $a,b$? Is there an intuitive explanation or we just need to accept it as a given fact?
74091
Problem: Show that if $(b,c)=1$, then for any integer $a$ we have $(a,bc)=(a,b)(a,c)$. [Hint: Prove that each member of the alleged equation divides the other.]  This is the way I proved it (without the hint): $(a,b)(a,c)=(a(a,b),c(a,b))=(a^2,ab,ac,bc)=(a^2,bc,a(b,c))=(a^2,a,bc)=((a^2,a),bc)=(a,bc)$  But how should I prove it using that hint or at least a similar proof.
81872
In a ring, I was trying to prove that for all $a$, $a0 = 0$.  But I found that this depended on a lemma, that is, for all $a$ and $b$, $a(-b) = -ab = (-a)b$.  I am wondering how to prove these directly from the definition of a ring.  Many thanks!
27649
It seems that given a statement $a = b$, that $a + c = b + c$ is assumed also to be true.  Why isn't this an axiom of arithmetic, like the commutative law or associative law?  Or is it a consequence of some other axiom of arithmetic?  Thanks!  Edit: I understand the intuitive meaning of equality. Answers that stated that $a = b$ means they are the same number or object make sense but what I'm asking is if there is an explicit law of replacement that allows us to make this intuitive truth a valid mathematical deduction. For example is there an axiom of Peano's Axioms or some other axiomatic system that allows for adding or multiplying both sides of an equation by the same number?   In all the texts I've come across I've never seen an axiom that states if $a = b$ then $a + c = b + c$. I have however seen if $a &lt; b$ then $a + c &lt; b + c$. In my view $&lt;$ and $=$ are similar so the absence of a definition for equality is strange.
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.
44952
Let $a$, $b$, and $c$ be positive real numbers. Prove that  $$\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}$$  Under what conditions does equality occur? That is, for what values of $a$, $b$, and $c$ are the two sides equal?
37035
Question:  If $ a^{2} + b^{2} = c^{2}$ and $c$ is even, prove that $a$ and $b$ are both even.  I am not quite sure how to prove this. My guess is proof by contradiction. Assume the contrary, that is, $a^{2} + b^{2} = c^{2}$ for $c$ even and at least one of $a,b$ odd.
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?
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: $$|x|=\begin{cases}x, &amp; \text{if $x\ge0$}\\-x,&amp;\text{if $x&lt;0$}\end{cases}$$