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24

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 $b = c$ ...


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 = C (because "=" is transitive and symmetric) Your teacher is right: if you prove that something true follows from A = B, i.e. A = ...


5

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.


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

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 ...


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

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.


3

My favorite : let $k$ be your ground field, and let $A = k[X_{ij}]_{1\leqslant i,j\leqslant n}$ be the ring of polynomials in $n^2$ indeterminates over $k$, and $K = Frac(A)$. Then put $M = (X_{ij})_{ij}\in M_n(A)$ the "generic matrix". For any $N=(a_{ij})_{ij}\in M_n(k)$, there is a unique $k$-algebra morphism $\varphi_N:A\to k$ defined by ...


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

There are two common errors that these methods are preventing: \begin{align} 1 &< 2 &&\text{True.} \\ 0 \cdot 1 &< 0 \cdot 2 &&\text{Um...} \\ 0 &< 0 &&\text{False.} \end{align} \begin{align} 1 &= 2 &&\text{False.} \\ 0 \cdot 1 &= 0 \cdot 2 &&\text{Um...} \\ ...


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

An alternative solution $$ \begin{align} E(X^r) & = \int_0^{\infty} r x^{r-1} P(X > x) \:dx \\ &= \int_0^{\infty} r x^{r-1} \left[ \int_{y=x}^{\infty} f(y) \:dy \right] \:dx \\ &= \int_{y=0}^{\infty} f(y) \left[ \int_{x=0}^{y}r x^{r-1} \,dx \right]\:dy \\ &= \int_0^{\infty} y^r f(y)\,dy \end{align} $$


2

Unfortunately your high-school teacher (and some of the other answers) is wrong. It is false that proving something of the form "$A = B$" is always possible by starting from one side and algebraically manipulating it to get a sequence of equal expressions ending with the other side, not to say necessary. Not necessary Let us first deal with the false ...


2

In finite dimensions (which is the case for you), if you know one basis $\mathcal B$, then proving another set $\mathcal L$ is a basis is easiest if you prove that $\mathcal L$ and $\mathcal B$ have the same number of elements every element from $\mathcal B$ can be expressed as a linear combination of elements from $\mathbb L$.


2

Let $L$ be the line $y=-mx+c$, $m>0$. The line shifted right by $a$ has the equation $$\begin{align*} y &= -m(x-a)+c\\ y &= -mx +c+ma\\ y-ma &= -mx+c \end{align*}$$ Since $m>0$ and $a>0$, $ma>0$, so the graph is equivalent to $L$ shifted up by $ma$.


2

$$2r-2s-1=2r-2s-1-1+1=2r-2s-2+1=2(r-s-1)+1$$ You can subtract 1 and plus 1 to balance things up. Something like adding zero, hence you preserve the equality.


2

Take $$S_n=\sum_{k=1}^n\left|f_k\right|$$ and apply the monotone convergence theorem for $\left\{S_n\right\}_{n=1}^\infty$.


2

It's just a matter of cardinalities. A countable union of countable unions of closed sets is still a countable union of closed sets.


1

Here is a neat proof from Qiaochu Yuan's answer to this question: If $L$ is diagonalizable with eigenvalues $\lambda_1, \dots \lambda_n$, then it's clear that $(L - \lambda_1) \dots (L - \lambda_n) = 0$, which is the Cayley-Hamilton theorem for $L$. But the Cayley-Hamilton theorem is a "continuous" fact: for an $n \times n$ matrix it asserts that $n^2$ ...


1

I'm not too sure what the etiquette is on Math SE regarding answering old questions but I stumbled across this question and felt it would be good to drop an answer for future users that happen to ask the same question in the future. Anyways, I'm going to do the particular case when $\delta(G) = 3$ and show that there exists a path of at least length 3. This ...


1

If we denote by $e_1,\dotsc,e_n$ the standard basis of $K^n$ (assuming $A\in {\rm GL}_n(K)$), then we have $Ae_1=0$ and $Ae_{i+1}=i\cdot e_i$ for $i=1,\dotsc,n-1$. Hence $A^{i}e_{i}=0$ for $i=1,\dotsc,n$. Then $A^n e_i = 0$ for all $i=1,\dotsc,n$, i. e. $A^n=0$ since $e_1,\dotsc,e_n$ is a basis. Note, that we have $A^{n-1}e_n = (n-1)!\cdot e_1$, which is ...


1

Hint:$ (x-y)^2>0$.Can you take it from here?


1

After some guidance, I think I might have the answer to this: Let $f,g$ be continuous on $[a,b]$ and $f(a)>g(a)$ but $g(b)>f(b)$. Prove that $\exists c \in[a,b]:f(c)=g(c)$. Proof: Define $h(x)$ as follows: $$h(x) = g(x) - f(x)$$Since $h(x)$ is a composition of two continuous functions, it shows that $h(x)$ is also continuous. Also, since ...


1

Since the set you're given has $n+1$ elements, all you need to show is that it is linearly independent, because $\dim P_n=n+1$. If you consider $$ a_0\cdot 1+a_1(x-1)+a_2(x^2-x)+\dots+a_n(x^n-x^{n-1})=0 $$ you get the matrix $$ \begin{bmatrix} 1 & -1 & 0 & 0 & \dots & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 & \dots & 0 ...


1


1

You need to prove a little lemma: (1) Sum of evens is even. (2) Sum of odd number of odds is odd. Prove (1) by factoring out a $2$, and prove (2) by induction on the number of terms. Then we can prove what you want. $E(G)=\{v\in V(G):d(v)\text{ is even}\}$ $O(G)=\{v\in V(G):d(v)\text{ is odd}\}$ $\sum_{v\in V(G)} d(v) =\sum _{v\in E(G)}d(v)+ ...


1

Statement: $A \cap (B' \cap C') = (A \cap B') \cap (A \cap C')$ $A \cap (B' \cap C') \subset (A \cap B') \cap (A \cap C')$ $x \in A \cap (B' \cap C') \implies x \in A$ and $x \in B' \cap C' \implies x \in A, x \in B', x \in C' \implies x \in A \cap B'$ and $x \in A \cap C' \implies x \in (A \cap B') \cap (A \cap C')$ $(A \cap B') \cap (A \cap C') ...


1

By definition, the elements of $\left(\mathbb{Z}/n\mathbb{Z}\right)^{\times}$ are the units of the ring $R = \mathbb{Z}/n\mathbb{Z}$. To say that $a\in R$ is a unit is to say that there exists $x\in R$ such that $ax = 1$, but by definition of $R$ that just means that $ax\equiv 1\pmod{n}$. At this point you can write everything out in prime factors if you ...



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