# Tag Info

9

Your proof is fine. Another way to say the same thing is: $x\mathbin|n$ and $x\mathbin|n+1$ would imply that $x\mathbin|(n+1)-n=1$, contradicting with the premise that $x>1$.

7

You wrote I can't imagine many circumstances where mathematical notation and a formal proof system wouldn't suffice to convey the author's pertinent thoughts. The case is, mathematical proofs contain lot of hand-waving argument such as referring to symmetry or to consistent change of variables. Editors are eager to accept these arguments but theorem ...

7

I nominate Halmos. All of his writing is good, but you might look in particular at "Finite dimensional vector spaces".

6

I can’t think of any likely expansion of what you’ve written that makes it true. What is true is that \begin{align*}\binom{40}5&=\binom{17}5+\binom{17}4\binom{23}1+\binom{17}3\binom{23}2+\\&+\binom{17}2\binom{23}3+\binom{17}1\binom{23}4+\binom{23}5\;, \end{align*} and it has a very straightforward combinatorial proof. HINT: You’re trying to ...

6

You need to show that $$2-{1\over k}+{1\over(k+1)^2}\le2-{1\over k+1}$$ This is equivalent to showing $${1\over k+1}+{1\over(k+1)^2}\le{1\over k}$$ and, by clearing out the denominators, this is equivalent to showing $$k((k+1)+1)\le(k+1)^2$$ Can you take it from here?

6

By the rational root test, the possible rational roots of the polynomial have the form $$r=\frac{1}{k}$$ where $k$ is an integer (positive or negative) that divides $n$. Which of them are integer? When are they roots of the polynomial?

6

There is nothing wrong with your proof. However, the critical sentence, the one starting "Having...", is true, but is (at least to me) not convincing, or at least not as simple as it could be. It would be easier to subtract the two equations, giving $1 = x(t_2-t_1)$, deriving a contradiction.

5

If integer $x$ divides both $n,n+1$ $x$ must divide $n+1-n$ If general, $x$ must divide $a\cdot(n+1)+b\cdot n$ where $a,b$ are integers We have chosen $a=1,b=-1$

5

The question does not appear to be about Catalan numbers, but about binomial coefficients. I will write $C(n,k)$ in the more common form $\binom{n}{k}$. Then $$\binom{n}{r+1}=\frac{n!}{(r+1)!(n-r-1)!}=\frac{n!}{r!(n-r-1)!}\frac{1}{r+1}=\frac{n!}{r!(n-r)!}\frac{n-r}{r+1}=\binom{n}{r}\frac{n-r}{r}.$$ Remark: There are various ways to define the $n$-th Catalan ...

5

$1)$ $\lnot q$ as premise $2)$ $p$ or $\lnot s$ as premise $3)$ $p \rightarrow$ ($d\land q$) as premise $4)$ $e \rightarrow s$ as premise $5)$ $\quad p\quad$ Assumption $6)$ $\quad d \land q\;$ (3, 5) by modus ponens $7)$ $\quad q\;$ by simplification (6) $8)$ $\quad q \land \lnot q\;$ (1, 8) (And-introduction) $9)$ $\lnot p$, since assumption $p$ ...

5

If you believe the question, that you are looking for one possible value of $n$, then you should be able to find one by setting $f(1)=0$ which clearly gives you a linear equation for $n$. Why doesn't $f(-1)=0$ work? You need something like @egreg's approach to show there are no other possibilities.

5

Without induction: At least one of $n, n+1, n+2$ is even and exactly one is divisible by three. Hence $2 \ | \ n(n+1)(n+2)$ and $3 \ | \ n(n+1)(n+2)$. As $2$ and $3$ are co-prime, this means $6 \ | \ n(n+1)(n+2)$.

5

Hint: If $\displaystyle a+\frac1a$ is an integer then $\displaystyle \left(a+\frac1a\right)^2,\left(a+\frac1a\right)^3, \ldots$ are integers. Multiply the powers out and you should be able to see why $a^t+\dfrac1{a^t}$ is going to be an integer for positive integer $t$, using a combination of symmetry and induction.

4

$$3(k^2+k+2)=6\cdot\frac{k(k+1)}2+6$$ Now $k(k+1)$ being the product of two consecutive integers is even for integer $k$ Non-inductive method: $$n^3+5n=\underbrace{(n-1)n(n+1)}_{\text{Product of three consecutive integers }}+6n$$ See The product of n consecutive integers is divisible by n factorial

4

Realized how basic this proof is thanks to Winther's comment. So the definition is that summing 2 matrices means adding each $ij$th element to get the corresponding element in a new matrix. Or, if $\mathbf{C} = \mathbf{A} + \mathbf{B}$, where $\mathbf{A},\mathbf{B} \in \mathbb{R}^{m,n}$, then $\mathbf{C}_{ij} = \mathbf{A}_{ij} + \mathbf{B}_{ij}$ So, ...

4

You can answer this by contradiction. Assume it is rational and see what happens.

4

$$\bigcap_{n>0}\left(0,\frac 1n\right)$$is a subset of $(0,1)$. For each $x\in (0,1)$ there is some $n>0$ ($n>1/x$), such as $$\frac 1n < x$$ and for such $n$, $$x\notin \left(0,\frac 1n\right)$$so $$\bigcap_{n>0}\left(0,\frac 1n\right) = \emptyset$$

4

If you don't mind I'm gonna do some magic! $$2^{3n}=(2^3)^n=8^n=(7+1)^n$$ Now use binomial expansion to get \begin{align} (7+1)^n&=\binom{n}{0}7^n+\binom{n}{1}7^{n-1}+\cdots+\binom{n}{n-1}7+\binom{n}{n}7^0\\ &=\binom{n}{0}7^n+\binom{n}{1}7^{n-1}+\cdots+\binom{n}{n-1}7+1\\ (7+1)^n&=7k+1\\ \end{align} So We have ...

4

Hint: $2^{3n} - 1 = (2^3)^n - 1^n = (2^3-1)((2^3)^{n-1} + ... + 1)$

4

Hint: $4^{\frac{p-1}2}=(2^{2})^{\frac{p-1}2}=2^{p-1}$

4

You can prove before Lemma. Let $X$ be a finite set, and let $Y$ a subset of $X$. Then $Y$ is finite. Hint: If $f\colon X\to Z$ is a function from $X$ to $Z\subseteq X$, then it cannot be a bijection for any susbet $Z$ of $X$.

4

You are of course right. The issue is context, what are you expected to do? If it has already been proved in your course that all geometric series with "$|r|$" less than $1$ converge, and if a formula has been derived for the sum, then what you did is fine. But it is a peculiarly simple question to ask if one has that machinery in hand. So I am inclined to ...

4

This is bordering on too pedantic a marking. You could have said "by the definition of surjectivity, since $g:B\to C$ is surjective, for the given element $m\in C$ there exists an element $b\in B$ with $g(b)=m$". Similarly for the second "why?". Again, your answer is in fact very good, even perfect I would say. It clearly shows you understand the concepts ...

4

Would I be able to use the Fundamental Theorem of Arithmetic to show that $f$ is injective? This proves itself, right? If $2^a3^b=2^c3^d$, what does the FTA say? Then, could I say that $f(\Bbb N × \Bbb N)$ is an infinite set of the countable set $\Bbb N$, so then it is therefore countably infinite. Thus, there exists a bijection $g$ (after showing ...

4

Yes. You can argue that every element in $A$ is an element of $B$: For every element $a = 6t \in A$, where $t\in \mathbb Z$, we have $a = 3(2t)$, and $2t\in \mathbb Z$ since $t\in \mathbb Z$, so by definition, $a \in B$. Another way to summarize the above is to note that $A$ contains all and only even multiples of $3$, whereas $B$ contains all integer ...

4

Why? Because this has been the most efficient way to communicate proofs, as shown by the answers of Brian M Scott and Pew. But that may not always be so. Mathematical theorems in journals now often reach tens or hundreds of pages, for example, the famous proof of Wiles's Theorem, or the Classification of All Finite Simple Groups (itself an ongoing project ...

4

You've shown that if $x$ is odd, then $5x-7$ is even, but that is not what you were supposed to show. You were supposed to show that if $5x-7$ is even, then $x$ is odd. To see the problem with your method, let's use your method to prove this very similar statement: Let $x\in \Bbb Z$. If $4x-2$ is even, then $x$ is odd. Your method now goes like ...

4

Hint: $a_n = \left(1+\dfrac{1}{n}\right)^n\cdot \left(1+\dfrac{1}{n}\right)$

4

Your professors are not thinking much differently than you can. But the proofs they are supplying are (most of the time) the results of somebody thinking about the problem "intuitively", seeing why the proposition "has to" be true, then putting each step in that intuition into a justifiable statement. That last step is the one you are having trouble with. ...

4

I'm not sure the name of it but a simple proof would be: Since $a$ is a root of $f(x)$ we can write $f(x) = (x-a)g(x)$ for some polynomial $g(x)$. Then taking the derivative we get $$f'(x) = g(x) + (x-a)g'(x)$$ Now plugging in $a$ we get: $$0 =f'(a) = g(a) + (a-a)g'(a) = g(a)$$ So $a$ is a root of $g$ and we can write $g(x) = (x-a)h(x)$. Hence \$f(x) = ...

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