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43

We have the number $10^{20}+1$. Whenever we have something in this kind of form, we need to find an odd factor of the exponent. In this case $5 \mid 20$, so we can use $5$ as the factor. Now, we can say $10^{20}+1=(10^4)^5+1$. How does this help us? Well, if we say that $x=10^4$, we have the polynomial $x^5+1$. This polynomial has $-1$ as a zero, meaning ...


35

There's no conflict between your high school teacher's advice To prove equality of an equation; you start on one side and manipulate it algebraically until it is equal to the other side. and your professor's To prove a statement is true, you must not use what you are trying to prove. As in Siddarth Venu's answer, if you prove $a = c$ and $b = c$ ...


12

This is basically a long comment: The thing is that when we construct sequences in the manner you described ("Having picked $x_1,x_2,\ldots,x_n\in S$, pick $x_{k+1}$,...") is much simpler to grasp than trying to make every single statement formal. Basically we are using the following theorem (or some suitable variant): "Let $P(F)$ be a property about ...


10

I tend to disagree with Luiz Cordeiro's answer. I think the real reason why it appears that "no one cares" about this kind of thing is because the existing system of rewards and incentives in professional mathematics tend to confer an advantage upon one kind of mathematician at the expense of the other. Lets distinguish two kinds of mathematicians: Type ...


9

It is enough.. Consider this example: To prove: $a=b$ Proof: $$a=c$$ $$b=c$$ Since $a$ and $b$ are equal to the same thing, $a=b$. That is the exact technique you are using and it sure can be used.


6

Hint: You can probably finish the proof yourself if you use the fact that $g(n)=O(f(n))$ means that there is a $c_1$ and $N_1$ such that $g(n)\le c_1f(n)$ if $n\ge N_1$. (You wrongly interchanged the roles of $f$ and $g$.) Added: We have to prove two things, (i) $f(n)=O(f(n)+g(n))$ and (ii) $f(n)+g(n)=O(f(n))$. In proving (i), there is basically nothing to ...


5

Sure. Given two vectors $u,v \in \mathbb{R}^2$, we have $$ u \cdot v = ||u||||v|| \cos (\theta)$$ where $\theta$ is the angle between them. If we take $u,v,w$ to be three unit vectors, each at angle $120$ degrees / $\frac{2\pi}{3}$ radians from each other, then the dot product will always be $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$.


5

Consider the left coset space $G/H$, define $ K = (\alpha) $ (note that $|K| = 11$, so $K$ is a p-group) and let $ K $ act on $ G/H $ by left multiplication. The fixed point congruence theorem on the actions of nontrivial p-groups states that $ |G/H| \equiv \textrm{Fix}_K (G/H) \pmod{11}$, and since $ |G/H| = 10 $ we conclude that all of $ G/H $ is fixed by ...


5

short answer: equality is symmetric, implication is not (both are however transitive) longer answer: You are right: if you proof A = C and B = C for some terms/expressions/objects A,B,C then you are allowed to conclude that A = B (because "=" is transitive and symmetric) Your teacher is right: if you prove that something true follows from A = B, i.e. A = ...


5

To prove a statement is true, you must not use what you are trying to prove. The main problem is that you've misunderstood what the second teacher said (and possibly the meaning of the equals sign). What that second teacher is saying is do not take as prior facts what you're trying to prove. As one textbook puts it (Setek and Gallo, Fundamentals of ...


5

First observe that $ (x-y)^2 + (x-z)^2 + (y-z)^2 \geq 0 $. Next, expand the LHS to obtain: $ 2x^2 + 2y^2 + 2z^2 - 2xy - 2xz - 2yz \geq 0 $ Now you simply divide by two and add $xy + xz + yz$ to both sides.


5

The easiest here is to prove the contrapositive: if $A=\emptyset$ and $B=\emptyset$, then clearly $A\cup B=\emptyset$ and we're done.


4

There is no rule that one must show inclusion both ways in a set theoretical proof for the equality of two sets, it is just very often an easy way to do it. If you can do it without, by all means. I do question your step $X \bigcap (\bigcup_{\alpha \in I} A_\alpha)^c = X \bigcap (\bigcap_{\alpha \in I} A_\alpha^c)$ since you are using the fact that ...


4

If $T$ is such that for all $x\in \mathbb{R} $ we have $\sin(x+T) =\sin(x) $, then in particular, setting $x=0$, we have $$\sin T =0$$ So $T=k\pi$ with $k \in \mathbb{Z} $. To conclude, we then need to check that $\sin(x+2\pi) =\sin(x)$ using your formula above (and that $\pi$ is not a period, by plugging $x=-\pi/2$ for example).


4

Assume by contradiction that $A$ is unbounded. Let $a \in A$. Then for each $n$ there exists some $x_n$ such that $$d(x_n,a) >n$$ Now, $x_n$ has a converging subsequence $x_{k_n} \to b$. Now, for $\epsilon=1$, since $x_{k_n} \to b$ there exists some $N$ so that for all $n >N$ we have $$d(x_{k_n}, b) <1$$ Then, for all $n >N$ we have $$d(a,b) ...


4

Definition: Absolute continuity of a function of several variables $f(x,y,z)$ is absolutely continuous if : for every bounded region $R^3$ $f(x,y,z)$ is jointly continuous over $R^3$. $\frac {\partial f}{\partial x} $,$\frac {\partial f}{\partial y}$,$\frac {\partial f}{\partial y}$ exist and are measurable over X,Y,Z and XxYxZ and also: $\sum_{k} ...


4

Short answer: It's the wrong kind of proof Longer answer: In your proof, you cannot assume the conlcusion. If I assume that all squirrels have two tails, then it follows from my assumptions that all squirrels have two tails. This does not mean I proved that they do, of course.


4

Mathamatics textbooks are written for humans. As such the author doesn't try to be explicit about everything that happens but assumes that the reader can follow certain arguments on his own. On the other hand there's the field of automated theorem proving. In it proofs are formalized in a very explicit way because computers need explicit commands. That ...


4

In the simple case of $A>0$, it suffices to just look at the taylor series definition: $$1+A\leq 1+A+A^2/2!+A^3/3!+\cdots=e^A$$


4

The statement wishing to be proven is as mentioned equivalent to whether or not for $n>2$ and $k\in\Bbb N$ we have the following relation: $$k^n < (k+2)^n - (k+1)^n$$ This is false. Consider the validity of the statement for $k=6, n=3$. One has $6^3=216$ and $(6+2)^3 - (6+1)^3 = 512 - 343 = 169$ $$6^3 = 216> 169 = (6+2)^3 - (6+1)^3$$ Why ...


4

In hoping to see this more clearly, let's start with an example. Consider $\tau$ the euclidean topology on $\mathbb{R}$, that is to say the topology induced by the distant $d(x,y) := |x-y|$. The open interval $(1,2)$ is an element of the topology $\tau$ for any, i.e. $(1,2)$ is open. Therefore $(1,2) \in \tau$. More generally, any interval of the form ...


3

You have the correct approach. Just clean up a few details. You are trying to show that given any $\epsilon > 0$, there exists a partition $P_\epsilon$ such that $U(P_\epsilon,f) - L(P_\epsilon,f) < \epsilon$. First define $M$, $$M := \sup_{x \in [a,b]} |f(x)|.$$ Then it follows that for $i = 1, 2, \ldots, n$ we have $$M_i - m_i \leqslant 2M,$$ ...


3

Since $(e^x)' = e^x$ for all $x$, and $e^x > 1$ for $x > 0$, we have, for $x > 0$, $\begin{array}\\ e^x-1 &=\int_0^x e^t dt\\ &> \int_0^x 1 dt \qquad\text{since } e^x > 1 \text{ for } x > 0\\ &=x\\ \end{array} $


3

HINT: Show that the relation has only finitely many equivalence classes, say $C_1,\ldots,C_m$, and use the subscripts on the equivalence classes to define your function $f$.


3

I agree with @Siddharth Venu. If we have to prove a=b, At times, on proceeding from LHS you may end up in a stage(intermediate step) which you cannot solve any further and you continue to solve RHS to arrive at the same stage i.e, a=c and b=c so you can conclude that a=b these kind of equality problems often comes in trigonometry Many chapters like ...


3

Why do you need a special word? If you are proving $P\iff Q$ you can just talk about the statement $P\implies Q$ and its converse. If you think about it, the converse (and inverse, and contrapositive, and negation) all implicitly refer to the original statement. It is not just "the converse" it is "the converse of the statement." You can't get around ...


3

Your proof is just fine. :-) The rule "to show $A = B$, verify that $A \subseteq B$ and that $B \subseteq A$" is quite nice, since it often gives you a fruitful starting point for your proving efforts. However, it's not actually necessary to verify an equality using this rule. Any other sound method of proof can be used as well. In your case, you ...


3

Let $\epsilon>0$. We want to pick a $\delta$ such that $|x-(-1)|=|x+1|<\delta\implies|f(x)-f(-1)|=|f(x)|<\epsilon$. Note that $$ |f(x)|=\bigg|\frac{3x^{2}-2x-5}{2x+3}\bigg|=\bigg|\frac{(3x-5)(x+1)}{2x+3}\bigg|=\bigg|\frac{3x-5}{2x+3}\bigg||x+1|. $$ We are free top choose $\delta$ as we wish, and, in doing so, we will have a handle on making the ...


3

Say the largest element is $r$, It is obvious that $\frac{2+r}{2}$ is still en element of $S$, which is a rational number. What's more $\frac{2+r}{2} > r $, that is a contradiction.



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