# Tag Info

30

I would say something along the lines of: The proposed result is false. Here is a counterexample...

27

In these cases on my homework when I type it up, I just change the statement of the theorem to match what is true. Given that problem on my homework I would write $\textbf{Theorem:}$ It is not necessarily the case that the sum of two integers is odd. $\textbf{Proof:}$ Observe that $1+1=2$. Since $1$ is an integer and $2$ is not odd, we have proved the ...

24

If you want to have a "set containment" kind of intuition, you should probably do the opposite: thinking of $P \implies Q$ as $P \subseteq Q$. The intuition is that if you are in some situation where $P$ is true (i.e. "being contained in $P$") then $Q$ is also true for this situation ("contained in $Q$"). In particular, $Q$ may contain more than just $P$, ...

15

One should not worry too much about logic (negation, contrapositives), and instead think about the concrete problem, that is, just think about numbers. What can we discover about numbers such the sum of any two is divisible by the third? If $x,y,z$ are positive integers, with $x\lt y\lt z$, then it looks "hardest" for $z$ to divide $x+y$. Since $x+y\lt 2z$,...

15

Taking @SiongthyeGoh's comment, in the above picture, $AE=1$, $\angle BAE=x$, $\angle BAC=\beta$, and $\angle BAD=\alpha$. It follows that $DE=\sin(x+\alpha)$ and $CE=\sin(x+\beta)$. Applying the cosine theorem to $\triangle DCE$, the given expression is equal to $DC^2$, hence is a constant.

14

Prove or find a counterexample: the sum of two integers is odd. The above statement is not true and a counterexample can be easily found. It suffices to check the first few positive integers to discover that $1+1=2$. I think the idea is that since it's not a Proof you shouldn't precede the statement of the counterexample with the word "Proof" like ...

9

Since it seems you are thinking in terms of subsets, It is like saying that if $P\supseteq Q$ then $Q^c\supseteq P^c$ (in the sense that if $P$ is the set of all things we know to be true as our hypothesis, then the entirety of $Q$ is among those things we know as a result to be true. On the other hand if $Q^c$ is the set of all things we know to be true, ...

8

$f(x,y)=3x^2+2y^2$ is a continuous function from $R^2$ to $R$ and $(-\infty,6)$ is an open subset of $R$, so $f^{-1}(-\infty,6)$ is an open subset of $R^2$.

8

Let $$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\color {blue}{2\cos(\alpha-\beta)\sin(x+\alpha)}\sin(x+\beta)$$ Now use $2\cos A \sin B = \sin (A+B) - \sin (A-B)$. $$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\left[\sin\left(\alpha-\beta +x+\alpha\right)-\sin\left(\alpha-\beta -x-\alpha\right)\right]\sin(x+\beta)$$ $$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\left[\sin\left(... 7 Write g(x,y)=3x^2+2y^2-6, and note that g^{-1}((-\infty,0)) is the inverse of an open subset, thus is open since g is continuous. 7 It is essentially a biconditional, also known as an if and only if. An "if and only if" statement goes both ways. That is, p\iff q means "if p is true then q is true" and "if q is true then p is true." The statement "p is sufficient for q" means "if p is true, then q is true." The statement "p is necessary for q" means that if we don'... 7 Hint: It essentialy boils down to showing that metric spaces are Hausdorff. Take x,y \in X with x \neq y. Then d(x,y) > 0. Can you check that$$B\left(x,\frac{d(x,y)}{2} \right) \cap B\left(y,\frac{d(x,y)}{2}\right) = \varnothing?$$You should convince yourself with a drawing. To prove it formally, take z in that intersection and get a ... 6 Cite the counterexample. Since 1 + 1 = 2, the sum of two arbitrary integers is not always odd. 6 Half-line proof, without using elements: A\in\mathcal P(A), hence A\in\mathcal P(B), which means A\subseteq B. 5 The "theorem" you are proving, in the specific case you give, is not "the sum of two odd integers is always odd." A "theorem", by definition, is a true (or at least provable) statement. So the "theorem" cannot be "the sum of two odd integers is always odd," because your argument shows that statement to be false. In fact what you are proving is "the sum of ... 5 Assume \displaystyle\lim_{n\to\infty}x_n=L and \displaystyle\lim_{n\to\infty}y_n=M. Then, (x_n) is bounded and thus there is a constant c>0 such that$$|x_n|\leq c,\quad\forall\ n\in\mathbb N.$$Furthermore, given \varepsilon>0 there is n_0\in\mathbb N such that$$n>n_0\quad\Rightarrow\quad |x_n-L|<\frac{\varepsilon}{2|M|},\quad|y_n-...

5

This is wrong. $6$ divides the sum of the digits of $15$, but $6$ does not divide $15$. $6$ divides $12$, but $6$ does not divide the sum of the digits of $12$. What is true is that $3$ divides $n$ if and only if $3$ divides the sum of the digits of $n$. $6$ divides $n$ if and only if $3$ divides the sum of the digits of $n$ and the last digit of $n$ is ...

4

I think you should add some text to help the reader. Here's my interpretation of an acceptable presentation: Suppose that for any two integers $a$ and $b$, $a+b$ is odd. Letting $a=b=1$, it follows that $a+b=1+1=2$ is odd, a contradiction. Edit: As pointed out in the comments, your presentation is not acceptable, since you have written "Theorem: the ...

4

It looks to me like the biggest step you might be missing is how to formulate the statement you want to prove, which is the complement of the (false) assertion you were given in the exercise. You certainly wouldn't write something like this in a paper (instead you would just express that you are showing a counterexample), but it might be helpful to your ...

4

Yes, you can sort of do as you did and use the fact that $$x\in A\implies \{x\}\in P(A)\stackrel{\text{hypothesis}}{\implies} \{x\}\in P(B)\implies x\in B$$ But you can also directly observe that $$A\subseteq B\stackrel{\text{by def.}}{\iff} A\in P(B)$$ and use $A\in P(A)$ to conclude.

4

The statement you are trying to prove is true, so you are unable to disprove it. Hint: The statement says that for every $x$, there exists some $y$. So all you need to do is find some particular $y$ for which the equality holds. Try to prove the statement by using very simple numbers for $y$....

4

It's overloaded notation, for sure. But I have never seen it mean anything other than the following: Let $f : X \to Y$. Then, for $A \subseteq X$, $f(A) := \{ f(x) : x \in A \}$. The reason there isn't much notice is probably because it's a very common convention. It may even be explicit at some point in some books on Real Analysis. It turns out that ...

4

HINT Use the formulas $\sin^2t=\frac {1-\cos2t} 2$ and $\cos2p+\cos2q=2\cos(p+q)\cos(p-q)$.

3

It would be as follows: This theorem or proposition is false because the following example satisfy the hypothesis of the theorem but it doesn't satisfy the conclusion. Example .......

3

Note that $$\left| \frac { (-1)^{ n }\cos \sqrt { n } }{ \sqrt [ 3 ]{ n } } \right| =\frac { \left| (-1)^{ n } \right| \left| \cos \sqrt { n } \right| }{ \left| \sqrt [ 3 ]{ n } \right| } <\frac { 1 }{ \left| \sqrt [ 3 ]{ n } \right| } <\varepsilon$$ so $$n>\frac { 1 }{ { \varepsilon }^{ 3 } }$$ and take { n }_{ \varepsilon }=\left\...

3

If $\;0< \sqrt {x^2+y^2}<b\;$ then $\;|y|<b \;$ so $\;|x^2y/(x^2+y^2)|=|x^2/(x^2+y^2)|\cdot |y|\leq |y|<b.$

3

You should get rid of $x+5$, as it is a function of $x$. In order to do this, we fix an arbitrary delta say 1 and take minimum from 1 and a new delta which will be obtained. More precisley we already know that $|x-5|<\delta$. Now put $\delta=1$ and $9<x+5<11$ is obtained. Let us go back to our limit. One may see that $4(x+5)(x-5)<4(x-5)\times ... 3 As$f$is continuous on$D$(which is compact), it is so uniformly. Let$\epsilon > 0$. Then, there is$\delta > 0$such that if$\|(x,y) - (x',y')\| < \delta$, then$|f(x,y) - f(x',y')| < \epsilon/(d-c)$. Suppose that$x,y \in [a,b]$are such that$|x - y| < \delta$. Then, for any$t \in [c,d]$,$\|(x,t) - (y,t) \|= |x - y| < \delta$, so ... 3 To be a bit more formal: If I am choosing to disprove proposition$P$is false, by exhibiting a counterexample, I would be proving. "The proposition$P$is false." And the proof would start from "If$P$, then for all integer (or whatever)$n$,$<$some statement about$n$that is implied by$P>$". Therefore, in particular, for$n=\$ my ...

3

I would suggest trying to present your example in a different manner though mostly correct. I also recommend presenting it in a manner of using exact language to avoid ambiguity and variables so as to mathematically represent the statement and proceed to use a Proof by contradiction where a Counterexample would be used. So to present your problem: ...

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