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6

By Spectral Theorem, $A$ is orthogonally similar to a diagonal matrix, i.e $$ P^{T}AP=\pmatrix{\mu_1 \\ & \ddots \\ && \mu_n} $$ where $\mu_i>0$ is eigenvalue of $A$, and $\space P^{T}P=P^{-1}P=I$. For any $v$, let $v=Pu$. Then $$ v^TAv=u^TP^TAPu=\sum_{k=1}^n\mu_ku_k^2>0 $$


4

I would write it as: Let $P(n)$ stand for the expression: $$\forall x\leq n(x\not\in A)$$ Then use the assumption that $A$ has no least element to prove that $P(1)$ and $P(n)\implies P(n+1)$. Thus, we've shown that $\forall x:x\not\in A$, which means $A$ is empty. That's essentially the same as your proof, but uses less set notation.


4

Usually, a proof by contradiction of the statement $p \implies q$ is when you assume that the opposite of the desired conclusion is true (i.e., assume the negation of $q$ is true), and follow a few logical implications until you reach a statement that somehow explicitly or implicitly contradicts an initial assumption from the statement $p$. Meanwhile, a ...


3

You won't be surprised to learn that in the last seventy-plus years since Tarski's book was first published in English, many other books have been appeared which will perhaps serve better as introductions to modern logic. And if you have downloaded my Teach Yourself Logic, you will have seen my "entry-level" suggestions on formal logic at the beginning of ...


3

i) Given $$y=\frac{3x^2+2y}{x^2+2}$$ cross multiplying we have, $$y(x^2+2)=3x^2+2y$$ $$\implies yx^2+2y=3x^2+2y$$ canceling $2y$ in both side we have, $$yx^2=3x^2$$ Since $x\neq 0 \implies x^2\neq0$ So dividing both side by x^2, we have $$y=3$$ ii) On contrary suppose that $x=0$ then we have $x^2=0$ so $$0\cdot y=2\cdot 0 +y$$ $$\implies y=0$$ Which is a ...


3

As has been pointed out in comments, we do not need to add a constant symbol for every element of $\mathbb{R}$. Let $L$ be the usual first-order language for ordered fields. Extend $L$ to $L'$ by adding a constant symbol $c$. Let $T$ be the set of all sentences of $L$ true in the reals, under the usual interpretation. Make a new theory $T'$ by adding to $T$ ...


3

Instead of compactness, you can try an ultrapower argument. Let $\cal U$ be a nonprincipal ultrafilter on $\Bbb N$. Then $\Bbb{R^N}/\cal U$ is elementarily equivalent to the reals. And it is not hard to show it has the wanted cardinality, and that it is non-standard.


2

You can do them more simply as: $P(AB)\leq P(A)=0$ hence $P(AB)=0$. But $P(A)P(B)=0$ so $P(AB)=P(A)P(B)$ implying $A,B$ are independent. $P(A\cup B)\geq P(A)=1$ hence $P(A\cup B)=1$. So $P(A)+P(B)-P(AB)=1\implies P(B)=P(AB)\implies P(A)P(B)=P(AB)$ so $A,B$ are independent. Suppose $D,D^c$ are independent. Then $P(DD^c)=P(D)P(D^c)\implies ...


2

(ii) Exists an $M$ such that for each $x$ real, $|f(x)| \leq M$. (iii) For each $x$ real there exists an $M$ such that $|f(x)| \leq M$. In the first case $M$ is unique for all $x$. In the second case $M$ depends on $x$. In this sense, we can restate (iii) as follows: (iii) For each $x$ real there exists an $M(x)$ such that $|f(x)| \leq M(x)$. The ...


2

It's all right, but in point b) you have mistake: need $\ge$, not $\le$.


2

No, for example: let $A = \{\frac 1 k \mid k \in \mathbb N\}$, $N = \mathbb N$. Then $$ \exists n \in N \forall x \in A \colon xn \ge 1 $$ is false, since for a fixed $n \in \mathbb N$ we have $\frac 1 {n+1} \in A$ and $n \cdot \frac 1 {n+1} < 1$. On the other hand $$ \forall x \in A \exists n \in N \colon nx \ge 1 $$ is true. (Given $x = \frac 1 k \in A$ ...


1

(I) (∀x∈R)(∃n∈N)(nx<1) (II)(∃n∈N)(∀x∈R)(nx≥1) how is the negation of (II) equal to (I)? The negation of (II) is not (I). The negation of (II) is $(\forall n\in N)(\exists x\in R)(nx<1)$


1

This is not a proof by contradiction, I suspect this is not a proof at all you showed that $\forall (n_0,m_0,u_0), G(n_0,m_0,u_0) $ admits an example which has no PNE's. This is not the same as saying that $G(n_0,m_0,u_0)$ has no PNE's. note This will be a direct proof if you knew that if for some example $G(n_0,m_0,u_0)$ has no PNE's then ...


1

Suppose (i) $\langle 3, 4,5\rangle = \langle 4, 3,5\rangle$ is true. Let's try to prove it then. According to the definiton, two sequences must be the same length, have the same items, and have their items in the same order to be equal. So we check: The number of terms in LHS: 3; in RHS: 3 — they are equal; CHECKED. The items of LHS sequence are 3, ...


1

Hint. Show that the pairs $a + cb, b$ and $a, b$ have the same linear combinations. That is, show that every linear combination of $a + cb$ and $b$ is a linear combination of $a$ and $b$. Conversely, show that every linear combination of $a$ and $b$ is a linear combination of $a + cb$ and $b$


1

I'd recommend you use the following standard symbols: $∂(A)$ is boundary of $A$. $\overline{A}$ is the closure of $A$. $A^c$ is the complement of $A$. $A^o$ is the interior of $A$. $A'$ is the limit point set of $A$. Boundary is the set whose points are in the closure but not in the interior. So it is defined as $$ ∂(A)= \overline{A} - A^o $$ Also ...


1

Since the OP has stated that contradiction is needed, here's another answer. Consider the statement: If $x\not=0$, then if $y=\frac{3x^2+2y}{x^2+2}$ then $y=3$. The hypothesis is $x\not=0$ and the conclusion is if $y=\frac{3x^2+2y}{x^2+2}$ then $y=3$ To prove this by contradiction, you need to assume the hypothesis and the negation of the conclusion. The ...


1

For ii), a better starting approach is to assume that $y=0$ and $x \neq 0$. You are trying to find a contradiction. Then for the left hand side you would have: $$ x^2 y = x^2 (0) = 0$$ On the right hand side: $$ 2x + y = 2x + 0 = 2x \neq 0 \text{$\qquad$ by assumption that $x \neq 0$}$$


1

I would probably do something more like: $\forall x, y \in \mathbb{R}\,\,\,\exists\,\,z, g : x = z + g, y = z - g$ Proof Let Z=$\frac{x+y}{2}$ and G=$\frac{x-y}{2} $ $$\frac{x+y}{2} + g$$ $$=\frac{x+y+2g}{2}$$ $$=\frac{x+y+2(\frac{x-y}{2})}{2}$$ $$=\frac{2x+y-y}{2}$$ $$=x$$ I would repeat this for $y$ and then explicitly state my conclusion.


1

I would add the verification that $x=z+g$, $y=z-g$, or at least a sentence like "we see this indeed holds". Other than that it is fine. In fact, you don't need to show the system, if you show the verfication.


1

@Mathcraze already provided the way forward. I thought that it might be instructive to see the use of tensor notation herein. So, here we go ... Let $U$ be the orthogonal matrix that diagonalizes $A$, such that $UU^T=U^TU=I$ and $(U^TAU)_{ij}=\lambda_i\delta_{ij}$, where $\lambda_i$ is the $i$'th eigenvalue of $A$ and $\delta_{ij}$ is the Kronecker Delta. ...


1

If $A$ is a real symmetric matrix, then you can form an orthonormal basis for $\mathbb{R}^n$ from its eigenvectors. What happens when you expand $v$ in this basis?


1

The difference between : $\exists x \forall y$ and $\forall y \exists x$ is clearly shown by this example regarding natural numbers : $\forall n \ \exists m \ (n < m)$ is clearly true in $\mathbb N$, while : $\exists m \ \forall n \ (n < m)$ is false in $\mathbb N$.


1

Property 5 states that: $$\color{blue}{u}\cdot (\color{red}{v}\times\color{green}{w})=(\color{blue}{u}\times\color{red}{v})\cdot\color{green}{w}$$ Applying property 5 to the L.H.S. we get: $\color{blue}{(a\times b)}\cdot (\color{red}{c}\times \color{green}{d}) = (\color{blue}{(a\times b)}\times \color{red}{c})\cdot \color{green}{d}$ Property 1 states ...



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