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28

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 $b = c$ ...


19

The problem is that you did not state that those are equivalences (usually denoted by $\iff$) between your lines. And you do not even need the equivalences, you only need the implicatiosn from the bottom to the top, so you should perhaps write your proof "upside down". The way your proof is presented right now makes it look like the top implies the bottom. ...


7

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.


7

You can always find a place from which to see at least half the faces. To see why, start by considering a polyhedron with central symmetry. Imagine a viewpoint from which you don't see any lines as points or faces as lines (i.e. general position) and far enough away so that you can see all the faces whose normal points into your side of the half plane ...


7

HINT: $Z_{16}$ has an element of additive order $16$. Does $\Bbb Z_4\times\Bbb Z_4$? Added: You don’t want to confine your attention to ‘big’ properties like commutativity or being an integral domain; often the differences are only to be found at a more detailed level.


6

Your proof doesn't work because concluding a true statement is irrelevant. A proof by contradiction works because a true premise only yields true results, therefore if you get a false result your premise had to be false. It doesn't work the other way. Both true and false premises can yield true results so getting a true result yields nothing. Consider ...


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

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

You have the right idea: the quotient ring is a domain but isn't a field. It might be helpful to note that the quotient is isomorphic to $\mathbb{Z}[i]$, and then show that $2$ (for instance) isn't invertible.


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.


5

You don't need to use FLT. \begin{align} 3^{1974}+5^{1974}&\equiv (3^3)^{658} + (5^2)^{987} \pmod{13}\\ &\equiv (27)^{658} + (25)^{987} \pmod{13}\\ &\equiv 1^{658} + (-1)^{987} \pmod{13}\\ &\equiv 1 -1 \pmod{13}\\ &\equiv 0 \pmod{13}\\ \end{align}


5

In general, if $f: A \to B$ is a homomorphism of groups and $a\in A$, then $ord(f(a)) \le ord(a)$ because $a^n=1$ implies $f(a)^n=f(a^n)=f(1)=1$. When $f$ is an isomorphism with inverse $g$, we get $$ ord(a) = ord(gf(a)) \le ord(f(a)) \le ord(a) $$ and so $ord(a)=ord(f(a))$.


4

Your sum is the real part of $$\sum_{k=0}^{n-1}e^{i(2\pi k/n+\phi)}=e^{i\phi}\sum_{k=0}^{n-1}\left(e^{i2\pi /n}\right)^k=e^{i\phi}\frac{e^{in2\pi/n}-1}{e^{i2\pi/n}-1}=0.$$


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

Let $[\frac{x}{n}]=q$ then $q\leqslant \frac{x}{n}< q+1$. Then $nq\leqslant x < n(q+1)$. Hence $nq\leqslant [x] < n(q+1)$ and $q\leqslant [\frac{[x]}{n}]< q+1$. Thus $[\frac{[x]}{n}]=q$


4

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 ...


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

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

The many answers here correctly point out your error - you are working backwards. You seem to have trouble understanding those answers. Take comfort in the fact that you're not alone. Many students struggle with this problem. Here's a critique of your argument that may help you (it's essentially a rewording of @flawr 's accepted answer). The first line ...


4

If $x^2 -y^2=1$, then $(x-y) = \frac{1}{x+y} $, since $x$ and $y$ are positive integers $0<\frac{1}{x+y}<\frac{1}{2}$. But since x and y are integers so does $x-y$. Which gives us a contradiction.


4

Here is another take. $x^2-y^2=1$ implies $x^2=y^2+1$. $x^2=y^2+1 > y^2$ implies $x>y$, that is, $x\ge y+1$. But then $x^2 \ge (y+1)^2 = y^2+2y+1 \ge y^2+2+1 > y^2+1$, since $y \ge 1$. Thus, $x^2 > y^2+1$, contradiction. Another, more succinct way to express this is: $$ y^2 < y^2+1 < (y+1)^2 $$ Thus, $y^2+1$ is strictly between two ...


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

Let $B := U \setminus A$. Case I: Assume $x \in B' \setminus B$. \begin{align*} \implies & x \in A \\ \text{Let } & C = \{x\} \\ \implies & \left( C \setminus A \right) = \varnothing \ne C = \left( C \cap B' \right) \end{align*} Case II: Assume $y \in B \setminus B'$. \begin{align*} \implies & y \notin A \\ \text{Let } & ...


3

Hint: $(3x)!$ is a product including $3$, $6$, $9$, $\ldots$, $3 (x-2)$, $3 (x-1)$, $3 x$. You can pull at least one factor of $3$ out of each one of these and there are $x$ of them.


3

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


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

Assume there is one. Add 1. QED.


3

Since $n$ is a natural number we can divide $x$ (with remainder) by $n$ in order to express $x = nb +r_1$ for some $b \in \mathbb{N}$ and $r_1 < n$. Now $\lfloor \frac{x}{n} \rfloor = \lfloor \frac{nb+r}{n} \rfloor = \lfloor b + \frac{r}{n} \rfloor = b$ since $r<n$. On the other hand we have that $\lfloor x \rfloor = bn + r_2$ for the $\textbf{same}$ b ...


3

The only way I can make sense of "tautological shape" is to suppose a formula $\psi$ is of the form: $$\psi = \alpha[\phi_1 / p_1, \dots,\phi_n/p_n]$$ where $/$ denotes substitution (adequately defined), for some first-order formulas $\phi_i$ and a propositional tautology $\alpha$ in the variables $p_i$. Now let $\mathcal M$ be a structure for $\psi$ and ...



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