In a previous post, I looked at the sound for the word tea and the letter T in various European languages and found that they are the same in several languages. Recently I noticed that the fruit "orange" and the color "orange" are described by the same word in English. In what other languages are they described by the same word?
This is true in Chinese (both Mandarin and Cantonese), Spanish and French, but not Dutch. In the Dutch language, the color is called oranje (coincidentally orange is the national color of the Netherlands because it is also the name of the Dutch Royal family, derived from the name of a French principality belonging to their ancestor Willem de Zwijger), while the fruit is called sinaasappel or appelsien. I wonder what other languages have different words for these two concepts as well?
Sunday, October 27, 2013
Saturday, October 26, 2013
Theorem of the alternative and logical equivalence
In the theory of linear equalities and inequalities, several fundamental results are formulated as a theorem of the alternative: two alternatives are given, and exactly one of the alternatives is satisfied at any one time. A well known example is Farkas' lemma (1894):
For a real matrix $A$ and real vector $b$, exactly one of the following conditions is true:
where we say a real vector is (non)negative if all its entries are (non)negative. Consider a theorem of the alternative given in the following form: given two alternatives $P$ or $Q$, only one of them is true at any time, no more and no less. The truth table for this conclusion (denoted as $Z$) is then:
This means that $Z$ is equal to $P\oplus Q$ which is equivalent to $P\Leftrightarrow \neg Q$ which is equivalent to $\neg P\Leftrightarrow Q$.
Thus the theorem
Exactly one of the two statements $P$ and $Q$ is true
can also be restated as:
P is true if and only if Q is false
or as:
P is false if and only if Q is true
Setting the 2 statements in Farkas' Lemma above as $P$ and $Q$, then
$P\Leftrightarrow \neg Q$ is written as
Let $A$ be a real matrix and let $b$ be a real vector. Then there exists a nonnegative real vector $x$ such that $Ax=b$ if and only if for each row vector $y$ either $yA$ is not a nonnegative vector or $yb$ is nonnegative.
Given that $P\Rightarrow Q$ is logically equivalent to $\neg P \vee Q$, this can be rewritten as
Let $A$ be a real matrix and let $b$ be a real vector. Then there exists a nonnegative real vector $x$ such that $Ax=b$ if and only if $yb$ is nonnegative for each row vector $y$ such that $yA$ is a nonnegative vector.
which is the form of Farkas' Lemma as stated in A. Schrijver, Theory of Linear and Integer Programming, Wiley, 1998.
For a real matrix $A$ and real vector $b$, exactly one of the following conditions is true:
- There exists a real nonnegative vector $x$ such that $Ax = b$
- There exists a real row vector y such that $yA$ is nonnegative and $yb$ is negative.
where we say a real vector is (non)negative if all its entries are (non)negative. Consider a theorem of the alternative given in the following form: given two alternatives $P$ or $Q$, only one of them is true at any time, no more and no less. The truth table for this conclusion (denoted as $Z$) is then:
| P | Q | Z |
|---|---|---|
| 0 | 0 | 0 |
| 1 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 1 | 0 |
This means that $Z$ is equal to $P\oplus Q$ which is equivalent to $P\Leftrightarrow \neg Q$ which is equivalent to $\neg P\Leftrightarrow Q$.
Thus the theorem
Exactly one of the two statements $P$ and $Q$ is true
can also be restated as:
P is true if and only if Q is false
or as:
P is false if and only if Q is true
Setting the 2 statements in Farkas' Lemma above as $P$ and $Q$, then
$P\Leftrightarrow \neg Q$ is written as
Let $A$ be a real matrix and let $b$ be a real vector. Then there exists a nonnegative real vector $x$ such that $Ax=b$ if and only if for each row vector $y$ either $yA$ is not a nonnegative vector or $yb$ is nonnegative.
Given that $P\Rightarrow Q$ is logically equivalent to $\neg P \vee Q$, this can be rewritten as
Let $A$ be a real matrix and let $b$ be a real vector. Then there exists a nonnegative real vector $x$ such that $Ax=b$ if and only if $yb$ is nonnegative for each row vector $y$ such that $yA$ is a nonnegative vector.
which is the form of Farkas' Lemma as stated in A. Schrijver, Theory of Linear and Integer Programming, Wiley, 1998.
Wednesday, October 16, 2013
Gravity
Over the weekend, my family and I went to see the movie "Gravity" in IMAX 3D after hearing all the buzz about it. Admittedly there are some scientific errors in the movie (my son was whispering to me during the movie why Clooney needs to detach himself from Bullock to save her), the movie was a thrilling ride and made you feel what it is like to be adrift in space. I have always been fascinated by spaceflight ever since I read the books "The right stuff" and "2001: a space odyssey" (I have also seen the movies, of course) as a boy. The movie also reminds me of a common misconception we have when we see footage of astronauts in space: that there is no gravity in space (hence the term "zero gravity"). We see that as soon as the spacecraft escapes Earth's atmosphere, things like pens start to float around in the capsule. While is it true that in deep space, far away from other celestial bodies, the effect of their gravitational pull can be minuscule, all manned spaceflights are still relatively close to Earth and will feel its gravitational pull.
Consider astronauts at the International Space Station (ISS). They experience a gravitational pull similar in strength to what we experience on the ground. Recall Sir Isaac Newton's celebrated equation describing the forces between two bodies:
$$ F = \frac{Gm_1m_2}{r^2}$$
where $G\approx 6.67384 \times 10^{-14}m^3g^{-1}s^{-2}$.
Using Newton's second law of motion $F=ma$, the acceleration an object is subjected to due to another object with mass $m$ and a distance $r$ away is $\frac{Gm}{r^2}$. To calculate the acceleration induced by the gravitational force you feel from the Earth, if we assume the Earth is a sphere with a spherically symmetric density, then $r$ is the distance between you and the center of the Earth, approximately 6.371 Mm. The mass of the Earth $M_e$ is approximately $5.972\times 10^{27}g$, resulting in a constant $ \frac{G M_e}{r^2} \approx 9.82 ms^{-2}$.
The ISS is about 0.37 Mm above the ground, increasing $r$ to be about 6.731 Mm. The resulting constant is then approximately $8.8$, a 10% reduction in weight for everyone on board. However, the astronauts do not feel this force as they are orbiting the Earth, as they are essentially in continuous free fall. In other words, without gravity, the spacecraft would move in a straight line, and the speed of the spacecraft is such that the acceleration needed to change the trajectory to an orbit around the Earth is exactly equal to the acceleration induced by the gravitational force. It is a nice exercise in calculus (which by the way was also invented by Isaac Newton through his introduction of Fluxions and Fluents) to show this.
Since everything is relative, we must also note that we (along with the Earth) are orbiting the Sun as well at a speed of over 100Mm/hr. The distance from the Earth to the Sun is approximately $1.496\times 10^{11}m$ and the mass of the Sun is approximately $1.989\times 10^{33}g$. Applying Newton's formula above, we get $ \frac{G M_s}{r^2} \approx 0.00593$, which is much smaller than the constant for Earth. Since we and the Earth are in the same orbit (and thus in free fall) around the Sun, we do not notice the effects of the Sun's gravitational pull.
When I told my wife that we are moving around the Sun at high speed, she quipped "So we don't have a sedentary lifestyle then."
Consider astronauts at the International Space Station (ISS). They experience a gravitational pull similar in strength to what we experience on the ground. Recall Sir Isaac Newton's celebrated equation describing the forces between two bodies:
$$ F = \frac{Gm_1m_2}{r^2}$$
where $G\approx 6.67384 \times 10^{-14}m^3g^{-1}s^{-2}$.
Using Newton's second law of motion $F=ma$, the acceleration an object is subjected to due to another object with mass $m$ and a distance $r$ away is $\frac{Gm}{r^2}$. To calculate the acceleration induced by the gravitational force you feel from the Earth, if we assume the Earth is a sphere with a spherically symmetric density, then $r$ is the distance between you and the center of the Earth, approximately 6.371 Mm. The mass of the Earth $M_e$ is approximately $5.972\times 10^{27}g$, resulting in a constant $ \frac{G M_e}{r^2} \approx 9.82 ms^{-2}$.
The ISS is about 0.37 Mm above the ground, increasing $r$ to be about 6.731 Mm. The resulting constant is then approximately $8.8$, a 10% reduction in weight for everyone on board. However, the astronauts do not feel this force as they are orbiting the Earth, as they are essentially in continuous free fall. In other words, without gravity, the spacecraft would move in a straight line, and the speed of the spacecraft is such that the acceleration needed to change the trajectory to an orbit around the Earth is exactly equal to the acceleration induced by the gravitational force. It is a nice exercise in calculus (which by the way was also invented by Isaac Newton through his introduction of Fluxions and Fluents) to show this.
Since everything is relative, we must also note that we (along with the Earth) are orbiting the Sun as well at a speed of over 100Mm/hr. The distance from the Earth to the Sun is approximately $1.496\times 10^{11}m$ and the mass of the Sun is approximately $1.989\times 10^{33}g$. Applying Newton's formula above, we get $ \frac{G M_s}{r^2} \approx 0.00593$, which is much smaller than the constant for Earth. Since we and the Earth are in the same orbit (and thus in free fall) around the Sun, we do not notice the effects of the Sun's gravitational pull.
When I told my wife that we are moving around the Sun at high speed, she quipped "So we don't have a sedentary lifestyle then."
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