# Tag Info

108

Definition. Let $n\in \mathbb{Z}$ be an integer. $n$ is called even if ... According to the definition only integers are even or odd. It is not something that you have to prove.

85

The original concept of even and odd is defined on integers. However, one can ask: Is there a natural extension to all real numbers? And if so, what would be the evenness or oddness of irrational numbers? Now, what would we demand of such an extension? Well, the most important demand is, of course, that integers are even under the extended definition if and ...

38

One useful approach to generalize odd and even numbers to the irrational numbers is to quantify the "amount of evenness" of a number. We will see that $12$ is "twice as even" as $6$. Then, instead of asking whether an irrational number is odd or even, ask how even it is. First, start with Patrick Da Silva's answer to the linked question. He explains the ...

13

You should be able to delineate the precise mathematical theorems that allow you to make each step in a proof. For example, if you have $(x,y) \in \mathbb{R}^2$ and you write: let $r,\theta$ satisfy $x = r\cos \theta,y=r\sin \theta$ with $r\geq 0$ and $2\pi > \theta \geq 0$, you are using a theorem that says that: Proposition. For all $x,y \in ... 12 Hint: Show that one of the numbers is a multiple of$5$. One way to do that: Write$x=5k+r$. 11 Initial comments: This is an excellent question in my opinion and is just what the proof-writing tag is for. Unfortunately, there are often many problems plaguing beginners when it comes to induction proofs: Why induction is a valid proof technique should be understood at the outset, and this is rarely the case. Less relevant in high school or undergrad, ... 10 I can relate. Throughout the years, many of my exams and homework assignments had deductions for not being rigorous enough. The trick to mathematical rigour is managing the leaps that you make from statement to statement. The size of the leaps is inversely proportional to the rigour. Let me mention some heuristics that tend to work well for me: If you ... 10 What you need first and foremost is not a proof but a definition. Normally, one will say something like: one calls an integer even if it is divisible by$2$, and odd otherwise. Then, being "even" and "odd" is a property of integers, and it makes no sense to ask it for other numbers. It also makes no sense to ask if a circle is even. Now, you might have ... 9 (a) Let$f(x_1)=f(x_2)$. Then,$g(f(x_1))=g(f(x_2))$, but since$g\circ f$is injective... (b)$g(Y)\supseteq (g\circ f)(X)=g(f(X))$. Hence, if$(g\circ f)(X)=Z$... 8 An irrational number only makes sense if we look at numbers beyond the rationals. In any extension of the rationals we will find the number$\frac 12$. And then any number is divisible by$2$. We could say, in this context, that every number is even. The problem with this is that the concept simply ceases to be useful when used in that way. So we don't use ... 8 Your cardinality argument is wrong since finitely many product of at most countable set is at most countable. (Especially the product of finite set and countable set is countable.) Moreover, there are nonismorphic groups$A$and$B$which satisfy$A\times B\cong A$. (e.g.$A=\Bbb{Q}^\Bbb{N}$,$B=\Bbb{Q}$) Despite of this, you can prove that$\Bbb{Q}\times ...

7

$x\bmod5=x\bmod5$ $(x+6)\bmod5=(x+1)\bmod5$ $(x+12)\bmod5=(x+2)\bmod5$ $(x+18)\bmod5=(x+3)\bmod5$ $(x+24)\bmod5=(x+4)\bmod5$ so if $x\ne5$, then $5$ must divide one of the five integers, and it can't be $5$ itself, whence it must be composite.

7

I think you will profit from re-reading the definition of orbit. After you've done that, consider $z \mapsto -z$ and prove that it constitutes an example.

7

Definitions are just shorthand. For example "$f$ is a function from $A$ to $B$" is shorthand, in set theory, for $$f\subseteq A\times B \land \left(\forall x(x\in A\implies\exists_1 y((x,y)\in f))\right)$$ And here, even $\exists_1$ is a shorthand. Definitions are a way to avoid writing the same thing over and over again. Another example: Saying "$p,q$ ...

6

A proof of a proposition is rigorous if it convinces the reader that the proposition is true beyond a reasonable doubt. In math, as in everything else, what constitutes "reasonable doubt" is flexible. It depends on the cultural context: What was rigorous to Euclid was not always rigorous to Hilbert. What was rigorous to Gauss was not always rigorous to ...

6

Hint $$\sqrt{2}x-\sqrt{x^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{x},x>0$$ it is easy to prove by derivative. so $$\sqrt{2}a-\sqrt{a^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{a}\tag{1}$$ $$\sqrt{2}b-\sqrt{b^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{b}\tag{2}$$ $$\sqrt{2}c-\sqrt{c^2+1}\ge\dfrac{\sqrt{2}}{2}\ln{c}\tag{3}$$ $(1)+(2)+(3)$ ...

6

HINT: Recall that there is a bijection between $\Bbb Z$ and $\Bbb{Z\times Z}$. Can you find a bijection $f\colon\Bbb{Z\times Z\to Z\times Z}$ with infinitely many infinite orbits?

6

Taking logs, we want: $$y\ln x= x\ln y \\ \implies \frac{\ln x}{x} = \frac{\ln y }{y}$$ So looking at the function $f(x)= \dfrac{\ln x}{x}$, we find it increases to $x=e$, peaks there and decreases toward $0$ so has two $x$ values corresponding to each $0<f(x)<1/e$. So the answer is that all reals $x >1, x\ne e$ have a corresponding distinct ...

5

It looks right. Another possible approach is the following one - since: $$f(x)=x^2+3x+2 = \left(x+\frac{3}{2}\right)^2-\frac{1}{4}$$ we have that $f(x)$ attains its minimum in $x=-\frac{3}{2}$ (the abscissa of the vertex) and for every $r>-\frac{1}{4}$ the equation $$f(x)=r$$ has two solutions, symmetric with respect to $x=-\frac{3}{2}$. It follows ...

5

I guess that the original "motivation" for the odd/even distinction in human culture was to know if a number of things (like apples or stones) could be divided between two people in a fair way without having to cut or break a remainder one. And also to known if things could be paired: three women and three men was fine, but one woman and two men meant ...

5

This result holds in the general case with $2m$ objects of total weight $4m$, with none of the weights exceeding $2m$. If a list P of positive integers sums to $n$ then P is called a partition of $n$. Let |P| denote the number of elements in P. Let S' be the set of partitions P' of $2n$ with |P'| = $n$ (i.e., each partition P' in S' has $n$ elements and ...

5

stars and bars - represent the r identical object as r stars ****** you can divide them into boxes by placing bars in between stars so **|*|*** represents six objects split into three boxes with 2 in the first box, 1 in the second, and three in the third. in general there will be $r$ stars , leaving $(r-1)$ positions to place the $(n-1)$ bars. The total ...

5

Look at $0^2\pmod{p}$ and $1^2\pmod{p}$. EDIT If you exclude $0$, from the list of quadratic residues as some people do, the proof is as follows. Assume that there are no consecutive quadratic residues. Note that $1$ and $4$ are quadratic residues for all $p \geq 7$. If $2$ or $3$ or $5$ are quadratic residues we are done. Hence, we can assume that $2$, ...

5

$$\frac{1}{n + 1} + \frac{1}{n(n + 1)} = \frac{n}{n(n + 1)} + \frac{1}{n(n + 1)} = \frac{n + 1}{n(n + 1)} = \frac{1}n$$

4

That's odd; you shouldn't need that. One can proceed as follows: Assuming that we know $$F^2_n-(-1)^n = F_{n+1}F_{n-1}$$ we can write \begin{align} F^2_{n+1}-(-1)^{n+1} & = F_{n+1} (F_n+F_{n-1}) + (-1)^n \\ & = F_{n+1}F_n + F_{n+1}F_{n-1} + (-1)^n \\ & = F_{n+1}F_n + F^2_n \\ & = (F_{n+1}+F_n)F_n ... 4 There is a whole family of examples similar to the proposition that n^3-n is divisible by 6 for each natural number n. Proof by induction isn’t hard, but it’s certainly unnecessarily complicated. 4 Since B = P^{-1} A P, \tag{1} B^T = P^T A^T (P^{-1})^T; \tag{2} Now PP^{-1} = I, \tag{3} whence (P^{-1})^T P^T = I \tag{4} as well; but (4) implies (P^{-1})^T = (P^T)^{-1}; \tag{5} thus (2) becomes B^T = P^T A^T (P^T)^{-1}, \tag{6} so A^T is similar to B^T, putting the whole proof together! QED. 4 If B is similar to A, then we writeB = P^{-1}AP$$for some invertible P. Then transposing everything, we get:$$B^T = (P^{-1}AP)^T = P^TA^T(P^{-1})^T = P^TA^T(P^T)^{-1}.$$Since P is invertible, P^T is also invertible. So B^T is similar to A^T, and the matrix who plays the role of P in the definition is P^T now. 4 The matrix I+A^TA is invertible because it is positive definite: if v\in\mathbb{R}^n is nonzero, then$$ v^T(I+A^TA)v=v^Tv+v^TA^TAv=|v|^2+|Av|^2\geq |v|^2>0.  In particular, all the eigenvalues are positive and their product must too be positive (i.e. nonzero).

4

Suppose that $f$ is not injective, then there are $x,y$ such that $y\neq x$ and $f(x)=f(y)$, then we have $g\circ f(x)=g(f(x))=g(f(y))=g\circ f(y)$ which means that $g\circ f$ is also not injective. then by contrapositive we get $g\circ f$ injective $\implies f$ injective as well.

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