Is the following a complete proof? It is from the book, $Prealgebra$ by R. Rusczyk, D. Patrick, and R. Boppana  Consider the sum   $$x + (-x) + (-(-x)).$$  That is, we are adding x, its negation -x, and the negation of -x. By associative property of addition, we can add these three in any order. If we start by adding the first two, we have $x + (-x) = 0$, so  $$x + (-x) + (-(-x)) = 0 + (-(-x)) = -(-x).$$  However, suppose we start by adding $(-x) + (-(-x))$ first. Since $(-(-x))$ is the negation of $-x$, we have $(-x) + (-(-x)) = 0.$ So, we find  $$x + (-x) + (-(-x)) = x + 0 = x.$$  We just showed that $x + (-x) + (-(-x))$ equals both $-(-x)$ and $x$, so we must have  $$-(-x) = x.$$
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This is my attempt based on some stuff I have been seeing around:  Let $y = -x$, then $-y = -(-x)$.  Now, lets sum $y + x = (-x) + x = 0$, then we have $y + x = 0$. If we had the additive inverse of $x$, i.e $-x$, to both sides, we obtain:  $$y + x + (-x) = 0 + (-x) \\ y + 0 = 0 + (-x) \\ y = -x\\ -y = x \\ -(-x) = x$$.  Is this proof correct just using the field axioms? I think that the key point is finding that $y = -x$.
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Question: For every polynomial $P(x)$ of degree at n we have $$\sum_{i=0}^{n+1}(-1)^i\binom{n+1}{i}P(i)=0$$  Well I was inducting on degree of the polynomial, for $n=0$ its true and I got a bit confused in the inductive step, actually I cant figure the inductive step.   Is this property (if true) a famous one?  Thank you for your help.
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I'm self studying abstract algebra. I see that in rings there's no requirement for a multiplicative inverse.  Is there something similar except with no requirement for an additive inverse. For example, all the non-negative rational numbers. Every number other than 0 has a multiplicative inverse, but no additive inverses. We have both the multiplicative and additive identities. Multiplication is still associative over addition.  Is there a name for such an algebraic structure, and has it been studied the way rings have?
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If $c|a$ and $c|b$ then $c|(ax+by)$ where $c,a,b$ are integers  Proof:  Suppose $c|a$ and $c|b$ then we can represent as the following:  $a=cx$ where $x$ is an integer  $b=cy$ where $y$ is an integer  Then adding both:  $a+b=cx+cy=c(x+y)$  therefore $c|a+b$  but how can i show $c|(ax+by)$ ?
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I was doing some problems for Olympiad training and encountered this: How would you prove that  $(a+b+c+d)-(a+c)(b+d)\geq 1$?  We are told that $0&lt;a,b,c,d&lt;1$ and the product $abcd=(1-a)(1-b)(1-c)(1-d)$.  Thanks!!
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I need to use $|a+b| \leq |a|+|b|$ to show that $||a|-|b|| \leq |a-b|$ .   I have tried to represent  $||a|-|b||$ as  $||a|+(-|b|)|$ , and then get $||a|+(-|b|)| \leq |a|+|-|b||$ , but that isn't leading me anywhere given $|a-b| \leq |a|+|b|$.
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I was trying to find a proof of $a\times0 = 0$ by myself (assuming commutativity, associativity, distributivity, etc) and I came up with $$ a+0=a(1) \implies 1 = \frac{a+0}{a} = \frac aa + \frac 0a = 1 + \frac 0a \implies \frac 0a = 0 \implies a(0)=0$$ Is there anything wrong with this proof, or does it assume anything that shouldn't be assumed?
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How do we prove that $$ \sum_{i=1}^{n}(x_i-\bar{x})^2\lt\sum_{i=1}^{n}(x_i-a)^2 $$  where $a$ is any value other than $\bar{x}$, the arithmetic mean.  My Attempt: $$ \sum_{i=1}^{n}(x_i^2-2x_i\bar{x}+\bar{x}^2)\lt\sum_{i=1}^{n}(x_i^2-2x_ia+a^2)\\\sum_{i=1}^{n}x_i^2-2\bar{x}\sum_{i=1}^{n}x_i+\sum_{i=1}^{n}\bar{x}^2\lt\sum_{i=1}^{n}x_i^2-2a\sum_{i=1}^{n}x_i+\sum_{i=1}^{n}a^2\\-2\bar{x}.n\bar{x}+n.\bar{x}^2\lt-2a.n\bar{x}+n.a^2\\-2\bar{x}^2+\bar{x}^2\lt-2a\bar{x}+a^2\\-\bar{x}^2\lt-2a\bar{x}+a^2 $$ But how do I proceed further or is there any better way ?
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Of course this is a particular case of the closure under addition and multiplication of algebraic numbers (which proof would also be appreciated), but I would prefer a simple transformation on the already known polynomial giving $a$ as a root in order to obtain a polynomial giving $ia$ as a root. Thank you in advance for your help.
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The exercise is: Prove that $-(-v)=v$ for every $v \in V$  Proof Suppose  $v \in V$ and $V$ is a vector space. Then $-(-v) \in V$ as result of the scalar multiplication property and $-(-v)=-(-1\cdot v)=-1 \cdot(-1\cdot v) =(-1 \cdot-1)\cdot v = 1 \cdot v = v $ The desired result $-(-v)=v$ holds.  The solution manual gives the proof:  Proof   I just wanted to make sure that the way I did the proof isn't missing anything?Thanks
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Theorem:   If $x$ is a positive integer and  $x^3$ is even, then $x$ is even.   My Proof by Contrapositive:  I. Assuming that $x$ is odd, then I will show that $x^3$ is odd.  II. $x$ is odd, so $x$ can be expressed as $2k+1$  III. $(2k+1)^3$  IV. $2k(4k^2 + 6k + 3) +1$  V. Since $2k(4k^2 + 6k + 3) + 1$ is an integer, we can conclude $x^3$ is odd.  Did I go about this correctly? Should step V say "$2$ multiplied by an integer and added by $1$ is odd therefore we can conclude $x^3$ is odd"?
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Is my proof below correct? What specific property of rationals did I exploit in my proof? It looks like the property I exploited is the following: Given any positive rational, I can always write it as sum of arbitrary number of positive rationals, whereas given any positive integer I cannot write it as a sum of arbitrary number of positive integers. Has it got to do with the fact that $\mathbb{Q}$ is a field?       Problem Prove that the additive groups $\mathbb{Z}$ and $\mathbb{Q}$ are not isomorphic.   Solution Let there exist an isomorphism between $\mathbb{Z}$ and $\mathbb{Q}$. Now consider the element $1_{\mathbb{Q}}$. We then have $\phi(1_{\mathbb{Q}}) = z \in \mathbb{Z}$. Since $\phi$ has to be a bijection, $z$ cannot be zero, since $\phi(0) = 0$.  Now consider the element $\left(\dfrac1{z+1}\right)_{\mathbb{Q}}$. We now have $$z = \phi(1_{\mathbb{Q}}) = \phi\left(\underbrace{\left(\dfrac1{z+1}\right)_{\mathbb{Q}} + \left(\dfrac1{z+1}\right)_{\mathbb{Q}} + \cdots + \left(\dfrac1{z+1}\right)_{\mathbb{Q}}}_{z+1 \text{ times }} \right) = (z+1) \phi\left(\left(\dfrac1{z+1}\right)_{\mathbb{Q}}\right)$$ However, there is no element in $y \in \mathbb{Z}$ such that $(z+1)y = z$.    First update  Actually I realize that I complicated it unnecessarily. Instead, we can do like this. Since $\phi$ is an isomorphism, we have $\phi(q_{\mathbb{Q}}) = 1_{\mathbb{Z}}$ for some $q \in Q$. However, $$\phi(q) = \phi(q/2+q/2) = 2\phi(q/2)$$ And there is no $y \in Z$, such that $2y=1$. Hence, $\phi(q/2)$ remains unmapped.    Thanks
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Can't quite finish this proof:     Prove that if $x$ and $y$ are not both $0$ then $ x^2 +xy +y^2&gt; 0$   $ x^2 +xy +y^2 +xy -xy&gt; 0$  $ (x +y)^2 -xy&gt; 0$ Without loss of generality define $x\geq y$  Firstly the case where $y&lt;0$ and $x&gt;0$ is trivial as both $ (x +y)^2$ and $-(xy) &gt;0$.   I want to say by transitivity  as each are >0 that there sum must be ${}&gt; 0$ can I do that?  The second trivial case is $y=0$ as we simply have $ x^2 &gt;0 $ where $x&gt;0$.  Lastly we have define $c= x+y$  since both $x$ and $y$ must have the same sign we know $|c|&gt;x$ since $x\geq y$ we know that $ x^2 \geq xy$ finally since  $ |c|^2 = c^2$ $ \forall c \in \mathbb{R} $  We have $ c^2 &gt;x^2 \geq xy$ hence $c^2 -xy &gt;0$  I feel like I have all the pieces but it doesn't feel finished how do I fix it?  Edit: I am really glad i asked this question love the variety of different good answers no idea what a discriminant is but im going to go look that up. I wanted to give the accepted answer to Edward as he actually answered my question but Anurag s proof was just so much better then my attempt. ^^
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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.
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I am currently working on this problem from Hardy's Course of Pure Mathematics and have gotten stuck near the end. I was wondering if someone could help me determine what to go next.  Question  If $a_1, a_2, ...,a_n$ are all positive and $S_n=a_1+a_2+...+a_n$ then:  $(1+a_1)(1+a_2)...(1+a_n) \le 1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!}$  My Attempt  Proof by induction:  I had first shown that it was true for $n=1$ and $n=2$.  Now suppose that it is true for n. Then it must also be true for $n+1$  $(1+a_1)(1+a_2)...(1+a_n)(1+a_{n+1}) \le 1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!} + \dfrac{(S_{n})^{n+1}}{(n+1)!}$  Define: $(1+a_1)(1+a_2)...(1+a_n)=x$ and $1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!}=y$  Then: $x+xa_{n+1} \le y+\dfrac{(S_{n})^{n+1}}{(n+1)!}$  By the induction assumption, we know that $x \le y$.  What I can't figure out  From this point, what I think the next natural step would be is to show that  $xa_{n+1} \le \dfrac{(S_{n})^{n+1}}{(n+1)!}$  I know this rearranges to:  $xa_{n+1} \le \dfrac{(S_{n})^{n}}{n!} \times \dfrac{S_n}{(n+1)} $  And feel there may be something I can do here, but haven't been able to figure anything out.
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I was wondering why do some people use redundant axioms in definitions?  If you just expand $(a+1)(b+1)=(a+1)b+a+1=ab+b+a+1$ $(a+1)(b+1)=a(b+1)+b+1=ab+a+b+1$. Hence, $ab+a+b+1=ab+b+a+1$, then cancel ab and 1. Then, you get it's commutative for free.   Why do we then assume it?   Also, is it even stronger. If we drop the condition that 1 is in R, then can we still deduce a+b=b+a? As clearly the argument just done falls apart if you haven't got 1 in R.
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here says a new operation addition and new operation multiplication  x ⊕ y = x + y − 1, x ⊗ y = x + y − xy on set Z,where the operations on the right hand side are ordinary addition and multiplication of integers. Prove that the set Z equipped with these two new operations ⊕, ⊗ is a ring. Does it have a multiplicative identity?  I tried to use the addition cancellation but I do not understand why or I got the wrong understanding for this question.   I assume the there is y and y' are 2 solution for equation  x ⊕ y = x + y − 1, therefore:   x ⊕ y = x + y − 1  x ⊕ y' = x + y − 1 and I get  x ⊕ y =  x ⊕ y' after the cancellation, then I get y=y', and it means the addictive is unique. So, can I say that the set Z equipped with new addiction x ⊕ y = x + y − 1? or I just go into a wrong way for solve this problem?  ask in advance, thank you