48
votes
How to prove the following number theory puzzle
Suppose your numbers are $\{a_1,a_2,a_3\}$. Then the set of combinations you consider is $$\left\{\sum \delta_ia_i\quad \text {where}\quad \delta_i\in \{0, \pm 1\}\right\}$$
A priori, there are $3^3=...
- 68.4k
39
votes
Accepted
Why can you not divide both sides of the equation, when working with exponential functions?
The issue happens when you jump from
$$\frac{1}{6} = \left (\frac{4}{15} \right )^{x}$$
to
$$\frac{1}{6} = x\ln\left (\frac{4}{15} \right ).$$
In this step you have taken the logarithm of the right ...
- 1,995
39
votes
Algebraically why must a single square root be done on all terms rather than individually?
Note that $\sqrt{x} + \sqrt{y} = \sqrt{x+y} \iff (\sqrt{x} + \sqrt{y})^2 = (\sqrt{x+y})^2 \iff (\sqrt{x} + \sqrt{y})^2 = x + y$.
In order to intuitively understand it, let's substitute
$$ a = \sqrt{x},...
- 1,423
39
votes
Accepted
Why dividing equation by absolute value gives bad result
The mistake turns out not to have anything to do with the absolute values directly. Dividing both sides by an expression is only valid when that expression is nonzero. So as soon as we divide by $|x+2|...
- 75.2k
37
votes
Answer $9^x = 4^x + 6^x$ to a 10th grader, who knows math until the equation of a straight line (just before calculus)
$$9^x = 4^x + 6^x$$
Dividing by $4^x$ , we get :
$$\left(\frac{9}{4}\right)^x = 1 + \left(\frac{3}{2}\right)^x$$
$$\left(\frac{3}{2}\right)^{2x} = 1 + \left(\frac{3}{2}\right)^{x}$$
Substituting $\...
- 481
26
votes
Accepted
How to prove that $\log_5(6)>\log_6(7)$?
By change of base, you want to show
$${\ln (6) \over \ln (5)}>{\ln (7) \over \ln (6)}\iff \ln (6)>\sqrt{\ln (7)\ln (5)}.$$
This follows by AM-GM inequality:
$$\ln(6)=\frac{1}{2}\ln(36)>\frac{...
- 12.5k
26
votes
Accepted
Show that the polynomial $P(x):=x^4-6x+6$ has no real roots .
Write
$$x^4-6x+6=(x^2-1)^2+2\left(x-\frac32\right)^2+\frac12$$
and then it is clear that $x^4-6x+6>0$ for all $x\in\mathbb{R}$.
- 10.9k
26
votes
Accepted
Typo in math brainteaser book?
The book is making a mistake. That entire second half is nonsense, and the formula is incorrect.
In particular, it’s impossible to determine the speed of the river since there is no unit of distance ...
- 5,003
24
votes
Proving a polynomial to be positive for all real values
If $x=-1$, it is trivial. Otherwise,
$$
\sum_{j=0}^{2n}(-x)^j=\frac{x^{2n+1}+1}{x+1},
$$
and:
if $x>-1$, the numerator and the denominator of that fraction are both greater than $0$;
if $x<-1$, ...
- 4,891
22
votes
Real world example of an equation with no solution?
Sure. You and I run a race, but you get a five second head start over me. We both run at the same constant speed of 10 feet per second. At what time will we cross paths during the race?
The distance ...
- 12.5k
22
votes
Accepted
Equation $a^a=b^b$
Write $b=ax$ for some $x>1$. Then $x=b/a$ so $x$ is rational.
Then taking logs, we get
$$a\log a=b\log b=ax\log(ax)\implies \log a=\frac{x\log x}{1-x}
\implies a=x^{\frac{x}{1-x}}, b=x^{\frac{1}{1-...
- 10.9k
22
votes
Can you solve two unknowns with one equation?
You can rearrange your equation to give $$x\cdot (3y-2)=0$$
The product of two numbers (real or rational) can be zero only if at least one of them is zero. This is a useful property which is ...
- 99.5k
19
votes
Accepted
How to prove $\frac{\sqrt{5}+1}{2}>\log_23$?
Note that$$\frac{\sqrt{5}+1}{2}>\log_23\Longleftrightarrow2^{\sqrt5}>\frac92$$
we have
$$2^{\sqrt5}>2^{\sqrt{\frac{121}{25}}}=2^{11/5}>\frac92$$
This is true, because
$$2^{11/5}>\frac92 ...
- 21k
18
votes
Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$
Remark: @achille hui posted a similar proof. But we got them independently.
The desired inequality is written as
$$3^{2\sqrt 2} < (2\sqrt 2)^3$$
or
$$2\sqrt 2\, \ln 3 < 3\ln (2\sqrt 2)$$
or
$$\...
- 36.8k
18
votes
Algebraically why must a single square root be done on all terms rather than individually?
An equation is a scale, the kind with two pans, that balances when what's in the one pan weighs the same as what's in the other pan. You can do anything you want to do to one pan, and the scale will ...
- 177k
18
votes
Accepted
How to prove the following number theory puzzle
Let $a,b,c$ the set of integers. The goal is to find $14$ ways to respectively get $1,2,\ldots,14$. Each of $a,b,c$ can be with sign $+$, $-$, or $0$ (latter meaning that it doesn't occur). This makes ...
- 64.8k
17
votes
Accepted
Why using compound interest formula gives (potentially) wrong answer in this instance
The Khan academy answer seems to be derived from assumptions about how financial institutions operate. The various assumptions may reflect real-life finance (but not the mathematical viewpoint) or the ...
- 95.9k
17
votes
Why do we sometimes lose solutions when solving equations?
One frequent cause of losing solutions is dividing by something that could be $0$. For example, you get an equation of the form
$$ a b = b c $$
and you jump to the conclusion $a = c$ without taking ...
- 441k
16
votes
Show that $\sqrt{2} + 2 \sqrt{3} + 2 \sqrt{5} + \sqrt{7} < 12$
Rewrite the LHS as
$$\frac43\sqrt{1+\frac18}+4\sqrt{1-\frac14}+4\sqrt{1+\frac14}+\frac83\sqrt{1-\frac1{64}}$$
and use the binomial series to derive upper bounds for each term. For the series ...
- 102k
16
votes
Accepted
Value of expression $(\alpha^{110}+\beta^{110})-(\alpha^{98}+\beta^{98})$ in given quadratic equation
Let $\omega$ and $\omega^2$ be the roots of $x^2 + x + 1 = 0$. Then the roots of the equation $x^2 + 3^{1/4}x + 3^{1/2} = 0$ are $\alpha = 3^{1/4}\omega$ and $\beta = 3^{1/4}\omega^2$.
Using the fact ...
- 2,262
16
votes
Show that the polynomial $P(x):=x^4-6x+6$ has no real roots .
Still another way :
If $x≤0$, then $x^4-6x≥0\thinspace .$ Therefore, $x>0\thinspace .$ Thus, using the AM-GM inequality you have :
$$x^3+\frac 2x+\frac 2x+\frac 2x=6≥4\sqrt [4]{8}$$
A contradiction ...
- 14.6k
16
votes
Accepted
Similarities in the digits of the powers of 2 and 5
This is a consequence of the fact that $10 = 2 \times 5$, and thus:
$$5^{-n} = \frac{1}{5^n} = \frac{2^n \times 5^n \times 10^{-n}}{5^n} = 2^n \times {10}^{-n}$$
IOW, a negative power of $5$ is a ...
- 13.9k
15
votes
Can this be solved this without sin and cos?
Comment expanded to answer per request.
Let $T$ be the area of trapezoid under the shaded area $S_1$. Notice
$$S_1 - S_2 = (S_1 + T) - (S_2 + T)$$
and $S_1 + T$ is the area of a right triangle with ...
- 121k
14
votes
Accepted
Prove that $\prod_{1\leq i,j\leq n}\frac{1+a_ia_j}{1-a_ia_j}\geq1$ for $n$ real numbers $a_i\in(-1,1)$
Found here on AoPS:
For $-1 < x < 1$ we have the Taylor series
$$
\ln (1+x) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} x^k \, ,
$$
which implies
$$
\ln(1+x) - \ln(1-x) = 2\sum_{k=1}^\infty \frac{...
- 109k
14
votes
Show that the polynomial $P(x):=x^4-6x+6$ has no real roots .
We can use for example
$$x^4-6x+6 =(x^2-1)^2+2x^2-6x+5 >0$$
indeed $(x^2-1)^2\ge 0$ and
$$36-4\cdot 2\cdot 5 =-4<0$$
- 152k
14
votes
Accepted
How can we solve the particular equation $16x^5-200x^3-200x^2+25x+30=0$ in closed form?
This polynomial $f$ has Galois group $F_5$, the Frobenius group of order 20. Since this group is solvable, then $f$ is solvable by radicals. Here are Magma commands showing this.
...
- 19k
14
votes
Accepted
If 3 numbers of the form $\frac{1}{q^2+1}$ have a mean of $\frac{1}{2}$, will one of them always be $\frac{1}{2}$?
Unfortunately, the postulated conjecture is not true:
Indeed, examples of admissible $q$-triples that disprove the statement are
$$
(\tfrac{3}{2}, \tfrac{2}{9},\tfrac{41}{23})\\
(\tfrac{9}{2}, \tfrac{...
- 3,640
14
votes
Accepted
Show that $x<\dfrac{\sqrt [8]{128}}{2}$
collect $u = x^2 + 1 $ as $ \; \; \; x^{15} u^2 - u + x = 0$ so
$$ x^2 + 1 = \frac{1 \pm \sqrt{1 - 4 x^{16} \;}}{2 x^{15} } $$
and you cannot have $4 x^{16} > 1.$
Also $4 x^{16} = 1$ ...
- 138k
14
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 ...
- 9,869
13
votes
Derivative of irrational function is undefined at zero, but it seems weird for me.
The power rule (or the definition) correctly calculates that $f'(0)=0$. So the question is, why don't we get this answer using the product rule?
If we look closely at the product rule, this is what it ...
- 75.2k
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