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7

$\qquad\qquad\qquad\qquad$ $\qquad\qquad\qquad\qquad\quad$ Geometric Explanation of the Binomial Theorem $\qquad\qquad\qquad\qquad\qquad$ $\qquad\qquad\qquad\qquad\qquad\qquad$ Proof that $~\displaystyle\sum_{k=1}^n(2k-1)=n^2$


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It is correct, however, you might see the way similarly as: $$a(ab)b=a^2b^2=e=(ab)^2=a(ba)b.$$


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I'm fond of Euclid's proof of the infinitude of primes: For any finite set $S=\{p_1, p_2,\dots, p_k\}$ of prime numbers, let $N=p_1\cdot p_2\cdot\cdots\cdot p_k+1$. Then $N$ isn't divisible by any prime in $S$. Hence it is divisible by some other prime. Hence the set $S$ does not include all primes. Thus there must be infinitely many primes.


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Cosines and Sines Around the Unit Circle Trigonometric Angle Sum and Difference


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It does not just mean that there is always a number that is smaller than the absolute difference between the two given real numbers. Rather it means that there is such a number of the form $1/10^n$, so that digits in the $n$th decimal places in the two numbers differ from each other. PS: For example, two numbers differing in the $5$th place and not before ...


5

Proof of Euler's Identity: $$e^{\pi{i}}+1=0$$ BTW, your question is more or less a copy of "Simple" beautiful math proof, so you might wanna check it out too. There's some great colorful stuff there, my answer being somewhere in the middle.


5

I tried to find problems from different areas. My five suggestions are. Sophomore' dream. The formula for the problem is: $$\begin{align}\int_0^1 x^{-x}\,dx &= \sum_{n=1}^\infty n^{-n}\end{align}$$ You can find facts about the problem and the proof of it at Sophomore's dream wikipedia article. Bretschneider's formula. This is an expression for the ...


5

What you have is nowhere near a proof. The definition of $X$ can be accepted, but it is not conveying any insight transgressing the verbal formulation of the problem. We have to construct a bijective map $$f:\quad {\mathbb N}_{\geq0}\to X,\qquad n\mapsto A_n\ .\tag{1}$$ This map produces for each $n\in{\mathbb N}$ a finite set $A_n\in X$, and each element ...


4

Comments on your proof. Define a set $ X=\{A\subseteq\mathbb{N}\mid \text{$A$ is finite} \}$. OK We can have a function $g_{n}: \mathbb{N} \rightarrow A_{n} $ << Problem: You have to define $A_n$ Say you wanted to say, $A_n$ is a finite subset of $\mathbb{N}$ defined by: $$A_n = \{n_1,n_2,... | ~ n_i \text{ occurs as exponent in } n = p_1^{n_1} ...


4

You didn't show the asked property, all you've said is if there is an element "a" with such property then this element must be zero. Here is a simple proof of it in terms of basic ring properties: $$b\cdot 0 = b(0 + 0)= b\cdot 0 + b\cdot 0$$ $$\implies -(b\cdot 0) + b\cdot 0 = (-(b\cdot 0) + b\cdot 0) + b\cdot 0$$ $$\implies 0 = 0 + b\cdot 0= b\cdot 0$$ ...


4

Complete the square: $$2x^2-4x+3=(\sqrt 2 x)^2-2\times(\sqrt2 x)\times\frac2{\sqrt2}+\left(\frac2{\sqrt2}\right)^2-\left(\frac2{\sqrt2}\right)^2+3\\=\left(\sqrt 2 x-\left(\frac2{\sqrt2}\right)\right)^2+1$$


4

Here is an approach which does not require knowing anything about the square root function. $a^2 < b^2 \implies b^2 - a^2 > 0 \implies (b-a)(b+a) > 0$ which implies that $b-a$ and $b+a$ are nonzero and have the same sign. Therefore one of the following is true: $b > a$ and $b > -a$, so $b > |a|$, so $|b| > |a|$ $b < a$ and $b < ...


3

Proofs should follow the same rules as any other kind of writing. If the text naturally forms paragraphs then that's how it should be written. Now, usually if you have five pages of text forming a single paragraph that would be poor style, but I suppose it's possible that you could have a five-page-long paragraph if, say, one of the sentences of that ...


3

The rudimentary differential equation proof of Euler's formula in the complex plane $e^{i \pi}=-1$, where $i=\sqrt{-1}$. First, via $\frac{d}{d\theta}$, $$e^{i\theta}=f(\theta)+ig(\theta) \implies ie^{i\theta}=f^{\prime}(\theta)+ig^{\prime}(\theta)=if(\theta)-g(\theta).$$ Comparing real and imaginary parts, $f(\theta)=g^{\prime}(\theta)$ and ...


3

Calculus The proof that $\frac{22}{7} > \pi$. $$ \begin {align*} 0 &< \displaystyle\int_0^1 \frac {x^4 \left( 1 - x \right)^4}{1 + x^2} \, \mathrm{d}x \\&= \displaystyle\int_0^1 \frac {x^4 - 4x^5 + 6x^6 - 4x^7 + x^8}{1 + x^2} \, \mathrm{d}x \\&= \frac {22}{7} - \pi. \end {align*} $$ Geometry The Pythagorean Theorem. Algebra Proof ...


3

Here are a few visual proofs that $$\text{arctan}(1) + \text{arctan}(2) + \text{arctan}(3) = \pi$$ One by user KennyTM: More by user dldarek: I think the lattice nature of the proofs would look nice on a door.


3

$D = (-4)^2 - 4*3*2 < 0$, hence the polynomial $2x^2 - 4x + 3$ has no real roots, and so its graph does not intersect the horizontal axis. Thus, the value of this polynomial never change its sign, and, for, instance, when $x = 0$ it has the value $3 > 0$, so it always take positive values.


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This is a slight simplification of @Sami's solution: $$2x^2-4x+3=2(x^2-2x+1)+1=2(x-1)^2+1>0$$


3

$b*0=b*(1-1)=b*1-b*1=b-b=0$ Long: Let $b \in \mathbb R$. Consider $b*0$. By additive identity, and inverses, we have that $0=1 + (-1)=1-1$. By distribution, we have that $b*0=b*1-b*1$ This of course simplifies to $b-b$ and by the additive inverse property, $b-b=0$ forgive me if I'm assuming anything that your prof wouldn't allow, but I'm pretty sure I ...


2

As others have remarked, you seem to already assume that the set you want to be countable is countable by the way you label the finite sets. However, you can "lexicographically order" the finite sets of natural numbers in many ways, for example the following: If $A$ and $B$ are finite sets of natural numbers, say that $A <B$ if $|A| < |B|$. If $A$ and ...


2

The proof for the irrationality of $\sqrt{2}$ is pretty simple and satisfying, I think. It's a very easy result to achieve, but the proof is very elegant and has some nice symmetry. Assume $\sqrt{2} = \frac{p}{q}$ with p and q relatively prime (totally simplified). $2q^2 = p^2$ $p^2$ is even the square of an odd number is odd, so $p$ must be even. Let ...


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The classification of finite simple groups -- so there would finally be a single reference that could be given for this important result. ;)


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A still image from the top-voted entry at http://mathoverflow.net/questions/8846/proofs-without-words along with the equation it proves, $1+2+\cdots+(n-1)={n\choose2}$, could be good. (Note: the entry there was originally just a still. Personally I find the animation a little unpleasant, but that may just be me.) Added later: The original version of this ...


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Some suggestions: $1$. The proof for the Gaussian integral $$\int_{-\infty}^{\infty}e^{-x^2} \mathrm dx=\sqrt{\pi}$$ $2$. The proof for Euler's solution to the Basel problem $$\frac {\ \ \pi^2}6=\sum_{n=1}^{\infty}\frac 1{n^2}=\frac 1{1^2}+\frac 1{2^2}+\frac 1{3^2}+\frac 1{4^2}+\cdots+\frac 1{n^2}+\cdots$$ $3$. The proof for Wallis' product $$\frac \pi ...


2

The second sentence in you proof has a wrong preamble. It should read: There does not exist a function $f$ such that for $y_1, y_{2} \in \mathbb{N}$, if $y_{1} \neq y_{2} $, then $ f(y_{1}) \neq f(y_{2}) $. The last sentence is not clear at all. How is $k$ defined? I doubt you can turn that into a real proof. To achieve the result you can proceed by ...


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$$a^2-2ab+b^2=(a-b)^2=0\implies a-b=0\implies a=b$$


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Suppose that $P$ is a statement and $\sigma$ is the proof of proving $P$ directly. Now consider the proof which begins by assuming $\lnot P$, then repeating $\sigma$ to obtain $P$. Contradiction. The other direction is the impossible one: if you have a proof by contradiction for $P$, it might not be possible to obtain a direct proof for $P$.


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We can rewrite a direct proof, mechanically, as a proof by contradiction.



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