28
votes
Why do we sometimes lose solutions when solving equations?
If that is the case why do we sometimes lose solutions when solving equations?
We don't. If we take an equation and perform valid operations on it, we will never lose any solutions. If we look back ...
- 10k
21
votes
Why do we sometimes lose solutions when solving equations?
One frequent cause of losing solutions is dividing by something that could be $0$. For example, you get an equation of the form
$$ a b = b c $$
and you jump to the conclusion $a = c$ without taking ...
- 441k
16
votes
Why do we sometimes lose solutions when solving equations?
We lose solutions for the same reason that we lose keys, or lose our glasses: it is because we are not careful. If we are careful when solving equations, we will not lose solutions.
- 177k
10
votes
Why do we sometimes lose solutions when solving equations?
TL;DR (short response)
$$ ``\text{operation"} \neq ``\text{function"}. $$
But if we remove references to functions from the question, the question is a good one.
Discussion
The problem with the ...
- 96k
5
votes
Why do we sometimes lose solutions when solving equations?
You are actually more wrong than other answers suggest.
Notwithstanding the fact that, as already commented, operations are not functions and your notation is problematic, if you proceed like so:
$x=y ...
4
votes
Accepted
Find the minimum value of $\frac{1}{x - y} + \frac{1}{y - z} + \frac{1}{x - z}$ for real numbers $x > y > z$ given $(x - y)(y - z)(x - z) = 17$.
Since $x > y > z$, It seems natural to me to write
$$x - z > y - z > 0$$
hence let $$a = y-z, \quad b = x-y, \quad a+b = x-z.$$
Then we wish to minimize
$$f(a,b) = \frac{1}{a} + \frac{1}{b}...
- 129k
4
votes
Accepted
How solve the problem $f(x+2)=f(x)+4x+4$ for any $x$
Given, $f(x+2)=f(x)+4x+4$
$$1. \quad{f(x+2+2)=f(x+2)+4 \cdot (x+2) +4\\
2. \quad f(x+2+4){=\color{green}{f(x+4)}+4 \cdot (x+4) +4\\=\color{green}{f(x+2)+4 \cdot (x+2)+4}+4 \cdot (x+4) +4\\=f(x+2)+4\...
- 1,484
4
votes
Accepted
Find all the polynomials $f$ that satisfy $f(x^2)+f(x)f(x+1)=0$.
Zero-set
You have correctly identified two transformations which send roots of $f$ to roots of $f$. Explicitly, setting $u(z)=z^2$ and $v(z)=(z-1)^2$, we have $u(S)\subset S$ and $v(S)\subset S$. Let'...
- 3,376
4
votes
Accepted
How to prove $(1+\sqrt{2})^n$ is an irrational number for every $n \in \mathbb{N}$?
In your inductive step, assume that $\exists k, m > 0 : (1+\sqrt{2})^n = k + m\sqrt2$. Then:
$$(1+\sqrt{2})^{n+1}$$
$$= (k + m\sqrt2)(1 + \sqrt 2)$$
$$= k + k\sqrt 2 + m\sqrt2 + 2m$$
$$= (k + 2m) +...
- 14k
3
votes
Accepted
Inequalities for the solution of $x = (x-a) e^{x+a}$.
Trying to solve for $x$ the equation $$x = (x-a)\, e^{(x+a)}$$ means that you are looking for the inverse of
$$a=W\left(-x\,e^{-2 x}\right)+x \quad \text{for} \quad x>0\quad \text{and} \quad a>0 ...
- 256k
3
votes
Accepted
On the board there are numbers
Consider any four arbitrary positive numbers $\alpha, \beta, \gamma, \delta$. The product $P$ of these numbers is
$$P = \alpha \beta \gamma \delta$$
Suppose two arbitrary positive numbers $m,n \in \{\...
- 106
3
votes
Accepted
Is there a short proof that $-(-n) = n$, when $n$ is a positive integer?
This is a good question.
You are showing that $-n=(-1)n$, which is interesting in itself but not needed here.
As you said, $-a$ is the number that added to $a$ gives you $0$. So the original equality $...
- 202k
3
votes
Accepted
Why won't factoring lead to the same result as expanding when simplifying $(x+y)^2 - y^2 - x^2 + 2xy$?
$$(x+y)^2 \color{green}{- y^2 - x^2} + 2xy=(x+y)^2 \color{red}{-(y+x)(y-x)} + 2xy \tag 1$$
It is a wrong claim,
$$RHS{=(x+y)^2 - (y+x)(y-x) + 2xy\\
=(x+y)^2 - (y^2-x^2) + 2xy\\
=(x+y)^2 - y^2+x^2 + ...
- 1,484
3
votes
How solve the problem $f(x+2)=f(x)+4x+4$ for any $x$
Given that $f(x+2)-f(x)=4x+4$
here you can make a telescopic sum
$$\require{cancel}\cancel{f(x+2)}- f(x)=4x+4$$
$$\require{cancel}\cancel{f(x+4)}-\require{cancel}\cancel{f(x+2)}=4x+12$$
$$\cancel{f(x+...
- 919
3
votes
Accepted
Help me find a mistake in showing $ \sqrt[4]\frac{7-\cos 4x}{2} > -2\sin x $
When dealing with an inequality of the form
$$\sqrt[2n]{f(x)} > g(x),$$
you must take into account two possibilities. Either $g(x) < 0$, in which case it is sufficient to have $f(x)\geq 0$ for ...
- 7,332
3
votes
How to prove $(1+\sqrt{2})^n$ is an irrational number for every $n \in \mathbb{N}$?
Here's a very different approach that doesn't necessarily use induction.
Show that $ ( 1 + \sqrt{2})^n + (1 - \sqrt{2})^n$ is an integer $A$.
If we absolutely wanted, we can show this via induction.
...
- 66.9k
2
votes
How do you write "The sum of all the elements in set A"?
As long as we know that there is a way to add up elements of $A$, then writing $\sum_{a \in A} a$ is perfectly acceptable, as pointed out in the comments—but if you want to have an actual routine (or ...
- 381
2
votes
How solve the problem $f(x+2)=f(x)+4x+4$ for any $x$
Define $g(2x) = f(x)$, we find the difference sequence of $g$:
$$
0 \quad 12 \quad 32 \quad 60 \quad ... \\
12 \quad 20 \quad 28 \quad ... \\
8 \quad 8 \quad ... \\\
0 \quad ...
$$
Observe that the $4$...
- 1,426
2
votes
How do we prove that $(x^2 + y^2)^2 + x^2 + y^2 < 1$ is a disk?
Let $D = x^2 + y^2$ (so that $d = \sqrt{D}$ is the length of the vector $(x, y)$). Then the constraint is that:
$$
D^2 + D < 1
$$
But as $D$ is non-negative, $D^2+D$ is a monotone increasing ...
- 46.1k
2
votes
How do we prove that $(x^2 + y^2)^2 + x^2 + y^2 < 1$ is a disk?
If you let $x = r\cos(\theta)$ and $y = r\sin(\theta)$ where $r\geq 0$ and $\theta\in[0,2\pi)$, then it results that:
\begin{align*}
(x^{2} + y^{2})^{2} + x^{2} + y^{2} < 1 & \...
- 15k
2
votes
Accepted
Show $\sum_{i=0}^n{i\frac{{n \choose i}i!n(2n-1-i)!}{(2n)!}}=\frac{n}{n+1}$
By noting that
\begin{align*}
i \frac{\binom{n}{i} i! n (2n-1-i)!}{(2n)!} \cdot \frac{n+1}{n}
&= i \frac{(n+1)! (2n-1-i)!}{(n-i)!(2n)!} \\
&= \frac{\binom{i}{1}\binom{2n-1-i}{n-1}}{\binom{2n}{...
- 161k
2
votes
Show $\sum_{i=0}^n{i\frac{{n \choose i}i!n(2n-1-i)!}{(2n)!}}=\frac{n}{n+1}$
Supposing we start from
$$\sum_{q=0}^n {n\choose q} \frac{q}{q+1} {2n\choose q+1}^{-1}
= \frac{1}{n+1}.$$
The LHS is
$$n \sum_{q=1}^n {n-1\choose q-1}
\frac{1}{q+1} {2n\choose q+1}^{-1}.$$
Recall ...
- 60.5k
2
votes
How to determine argument from Euler’s form of complex numbers?
For a same specific complex number, changing $r$ will change the number. You can only maintain the number constant by changing its argument (by adding $2 \pi k, k \in \mathbb{Z}$)
Changing $r$ will ...
- 308
2
votes
Accepted
How to determine argument from Euler’s form of complex numbers?
If $z = a + bi$ then we have
$$\begin{eqnarray} a + bi & = & r e^{i \varphi} \\
& = & r \left(\cos \varphi + i \sin \varphi \right) \\
\implies a & = & r \cos \varphi \\
\mbox{...
- 22.5k
2
votes
Accepted
Function problem, pre-university level
Guide:
Since the quadratic function is even, the axis of symmetry is the $y$-axis, hence we know that $2a-3b=0$.
Also, we are told that the function is surjective, hence we know the minimum value is ...
- 148k
2
votes
Proving $\frac{a+b+c}{a^2b+b^2c+c^2a+9}\ge \frac{abc+6}{abc+27}$ for $a, b, c\ge 0$; $ab + bc + ca = 3$
Here is a proof.
If $abc = 0$, it is easy.
In the following, assume that $abc > 0$.
The desired inequality is written as
$$\frac{(a + b + c)(abc + 27)}{abc + 6} - 9 - (a^2b + b^2c + c^2a)
\ge 0$$
...
- 34.3k
2
votes
prove using induction or in any other way that for all natural number $n≥2$, $3^n>3n+1$
You want to prove that $3^{k+1}>3(k+1)+1$, so it's not very good to write this inequality from the beginning. You want to end with it. Start from the left hand side only:
$3^{k+1}=3\cdot 3^k>3(...
- 37.6k
1
vote
Finding the max of a function using $f(x)=L(x)-E(x)$ where $L(x)$ is linear and $E(x)$ is exponential
The question is ill-defined. Exponential functions have three degrees of freedom. Any exponential function can be written as $$E(x) = a(b)^x + c.$$
Note that $$ab^{x+c} + d = ab^xb^c +d = \tilde{a}b^x ...
- 1,511
1
vote
How do we prove that $(x^2 + y^2)^2 + x^2 + y^2 < 1$ is a disk?
Write it as $(x^2 + y^2 + 1/4)^2 < 5/4$. Since $x^2 + y^2 + 1/4 > 0$, this is equivalent to $x^2 + y^2 < \sqrt{5/4} - 1/4$.
- 441k
1
vote
Accepted
Volume of A Solid in 3-space consisting of all points (x,y,z) satisfying the Inequality.
Since you tagged this as pre-calc, I will answer that level. A circle is a special case of an ellipse (a = b = r). The area formula as a result will be
$$A = \pi \times r^2$$
The formula for the ...
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