# Tag Info

60

The product of three negative numbers is negative. So if $ab$, $ac$, and $bc$ are all negative, then $(ab)(ac)(bc)\lt0$. But $(ab)(ac)(bc)=a^2b^2c^2$ is the product of three squares, which are all positive.

12

HINT: One of the following is true about the integers: (all positive), (all negative), (one positive and two negative), or (two positive and one negative).

8

You don't need to find the unique solution, just to prove that it must be there somewhere. It is probably easiest to think of it as finding a zero of the function $f(x)=\ln(x) - \frac1x$. We know the function has at least one zero, because $f(1)=-1$ and $f(e)=1-\frac1e$ which is positive because $e>1$. Since $f$ is clearly continuous on $\mathbb R_+$, ...

8

This may be incredibly tedious to do, but you apply the sine sum formula to: $$\sin(48)=\sin(30+18)$$ Then, $\sin(18)$ and $\cos(18)$ can be computed by using the half angle formulas on $\sin(36)$ and $\cos(36)$, and noting that $\cos(36)=\frac{1}{2}\phi$, where $\phi=\frac{1+\sqrt{5}}{2}$, i.e. the golden ratio. This latter fact is proven here and here

7

Assume to the contrary, there exist $a$, $b$ and $c$ that are non-zero integers and none of the three products $ab$, $ac$ and $bc$ are positive. Things generally look OK up to this point. In particular, the statement above is a good assumption to make for the desired proof by contradiction. Now, I pick a = 1, b = -1 , c = 2 This is not OK: the ...

6

$x (1-x) = \frac 14 - (x - \frac 12)^2 \le \frac 14$ holds for all $x \in \mathbb R$. (Alternatively, use the GM-AM inequality, or visualize $y = x(1-x)$ as a parabola with vertex at $(\frac 12, \frac 14)$.) Then from $$a(1-b) \cdot b (1-c) \cdot c (1-a) = a(1-a) \cdot b (1-b) \cdot c (1-c) \le \frac 14 \cdot \frac 14 \cdot \frac 14$$ it follows that at ...

5

You ask for four real numbers satisfying that; so we can give $\sqrt 2,2\sqrt 2,3\sqrt 3,4\sqrt 3$ for which we get their mutual multiplication as: $4,3\sqrt 6,4\sqrt 6,6\sqrt 6,8\sqrt 6,36$. Let's see if we can construct an example by some arguments. If $a,b$ and $c$ are chosen as rational then $ab$, $ac$ and $bc$ are also rational and hence not ...

5

For $n=5$ you have $$n^2+2n=25+10 = 5 \times 7 = 35 \equiv 2 \pmod 3.$$ In general for $n= 6k-1$ you have $$(6k-1)^2+2(6k-1) = (6k-1)(6k+1) = 36k^2-1 \equiv 2 \pmod 3.$$

5

$$a^3+a=b^3+b\iff a^3-b^3=b-a$$ $$\iff (a-b)\left(a^2+ab+b^2\right)=b-a$$ If $a=b$, then we're done. For contradiction, assume $a\neq b$. Then $a-b\neq 0$ and we can divide both sides by $a-b$: $$a^2+ab+b^2=-1\iff 4a^2+4ab+4b^2=-4$$ $$\iff (2a+b)^2+3b^2=-4,$$ contradiction, because $(2a+b)^2+3b^2\ge 0$ for all $a,b\in\Bbb R$.

4

Consider this: $$4^{n+2} + 5^{2n+1} = 4\cdot 4^{n+1} + (21+4)\cdot 5^{2n-1} = 21\cdot5^{2n-1} + 4\cdot (4^{n+1}+5^{2n-1})$$

4

You can proceed with the inductive step as follows: Assume that $$21 \mid 4^{k+1}+5^{2k-1}$$which implies $$21\mid 25(4^{k+1}+5^{2k-1})$$ $$\Longrightarrow21\mid 25(4^{k+1})+5^{2k+1}$$ We can subtract a multiple of $21$ on the right side to obtain $$21\mid 25(4^{k+1})+5^{2k+1}-21(4^{k+1})$$ $$\Longrightarrow 21\mid4(4^{k+1})+5^{2k+1}$$ $$\Longrightarrow ... 4 A is singular iff there exists a nonzero vector X such that AX=0 iff 0 is an eigenvalue. No determinants required here, just the definition ;) 4 There are very many mistakes here: d_p is a distance on \Bbb R ^r, but you use it on \Bbb R when you write d_p (\phi ({\bf t}^n) , \phi ({\bf t})) the lines above the blue inequality are certainly false: the quantity between brackets may be negative, and if p=2 you get imaginary results; you may fix this with a modulus to get to the blue line, ... 4 This is not true, as \left(\begin{array}{c c}0 & 1\\ 1 & 0\end{array}\right) is invertible. 4 (The original question did not ask for a triangular matrix) There are (many) invertible matrices with a zero diagonal, for example consider the matrix$$ A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $$then$$ A^2 = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} $$so A is invertible with A^{-1} = A. Addendum: For the reformultated ... 4 A upper triangular matrix with 0's on its diagonal has its first column filled with 0's. Therefore its determinant is 0, which means it is not invertible. For a lower triangular matrix, the same holds with the last column. In case you're not familiar with determinants:When A is upper triangular, let's suppose A^{-1} exists. Then A^{-1}A=I. Let ... 4 So, you have seen for (a) that if a \ne 0, we have a single solution, namely x = a^{-1}(c-b). Hence, we are left with the case a = 0, that is with the equation$$ 0x + b = c \tag 1 As, in fields, we have 0x = 0 for any x - which can be seen as follows: \begin{align*} 0x &= (0+0)x & \text{0 is additive unit}\\ &= 0x + 0x ... 4 If they have the same parity, then their sum is divisible by two. but you have (3x+1)+(5x-2)=8x-1=2(4x-1)+1 odd. so they need to have different parities. 4 Hint: Use the definition. There is M such that f(x)<M for all x\in\Bbb R. What can you say about f(x+10)? 4 For n=2,{4\choose 2}=6> \frac{16}{3}$$Assume$${{2k}\choose{k}}>\frac{4^k}{k+1}$$Then$${{2k+2}\choose {k+1}}=\frac{(2k+2)(2k+1)}{(k+1)}{{2k}\choose k}>\frac{(2k+2)(2k+1)}{(k+1)}\frac{4^k}{k+1}={2(2k+1)}\frac{4^k}{k+1}=4k\frac{4^k}{k+1}+2\frac{4^k}{k+1}>4\frac{4^k}{k+1}=\frac{4^{k+1}}{k+1}$$... 3 Induction actually does work. Let a_n denote the LHS and b_n denote the RHS at the n^{\text{th}} step. Then$$ \frac{a_n}{a_{n-1}} = \frac{2n(2n-1)}{n^2} = \frac{4n-2}n, \quad \frac{b_n}{b_{n-1}} = \frac{4n}{n+1}. $$To figure out which factor is bigger, we consider their ratio :$$ \frac{\frac{4n-2}{n} }{ \frac{4n}{n+1}} = \frac{(4n-2)(n+1)}{4n^2} ...

3

Your proof is correct, but I feel that this could be proved by contradiction. Assume for contradiction $\exists x>0$ such that the equation $\sqrt{x+2}-\sqrt{x+1}=\sqrt{x+1}-\sqrt{x}$ is true. Then, \begin{align} \sqrt{x+2}-\sqrt{x+1}&=\sqrt{x+1}-\sqrt{x}\\ \frac{x+2-x-1}{\sqrt{x+2}+\sqrt{x+1}}&=\frac{x+1-x}{\sqrt{x +1}+\sqrt{x}}\\ ...

3

Your proof is correct, because each inequality you write is equivalent to the previous one (it should be noted, probably). Changing all $\ne$ into $=$ would make it a proof by contradiction, that's however unnecessary. In a different way, you could just swap terms and square, again changing inequalities into equivalent ones: \begin{gather} ...

3

Hint: Show that $g(x) := f(x)x^{-\alpha}$ is constant.

3

There is an easier way : assume that $A$ is diagonalizable, i.e. $A=PDP^{-1}$ with $D$ diagonal matrix and $P$ invertible matrix. Then write $$A+\alpha I_n=PDP^{-1}+\alpha PP^{-1}=P(D+\alpha I_n)P^{-1}$$ and as $D+\alpha I_n$ is a diagonal matrix you have your result.

3

If you are to derive this directly from the definition $\cos(x) \equiv \sum_{n=0}^\infty\frac{x^{2n}(-1)^{n}}{(2n)!}$ and $\sin(x) \equiv \sum_{n=0}^\infty\frac{x^{2n+1}(-1)^{n}}{(2n+1)!}$ then we can do this without having to perform the product by first using term-by-term differentiation to get $$\cos'(x) = -\sin(x),~~~~~\sin'(x) = \cos(x)$$ If we now ...

3

It's easy enough to make an injection either way: $C\subseteq (0,1)$ is the obvious injection one way, and for the other, there is a very simple bijection from $(0,1)$ to $(0.7,0.8)\subseteq C$.

3

Consider (as proposed by David Mitra in the comments) $$f(n+1)-f(n)=\frac{f(n+1)-f(n)}{(n+1)-n}=f'(n+\theta_n),\qquad \theta_n\in(0,1)$$ Now we know that the limits on both sides exist by the assumptions of the statement. The limit on the left is $α-α=0$, the limit on the right $β$ as $c_n=n+\theta_n\to\infty$.

3

Say $p$ does not divide $k$. Then gcd$(p,k)=1$. It follows that we can find integers $a,b$ with $$ap+bk=1$$ Multiplying by $l$ yields $$apl+bkl=l$$ As $p$ divides both terms on the left it divides their sum. Hence $p|l$.

3

HINT : Instead of mod 2, considering in mod 3 helps.

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