Dichotomy paradox
Exercise is impossible.
Because a moving object must reach its halfway point before reaching its destination, if the space is assumed to be infinitely separable, then the limited distance contains an infinite number of points, so the moving object will pass through an infinite number of points in a limited time.
It should have been put forward by Zhuangzi in "The World in Zhuangzi" at the earliest: "A hammer of one foot, take half of it every day, and it will last forever."
Achilles paradox
Achilles is a hero who is good at running in ancient Greek mythology. In his race with the tortoise, the tortoise ran ahead and he chased after it, but he couldn't catch up with it. Because in the competition, the pursuer must first reach the starting point of the pursued. When Achilles reached the tortoise's position at a certain moment, the tortoise had moved forward. When Achilles reached the turtle's position again, the turtle ran a little further; ..... So, no matter where Achilles goes, the tortoise will be in front of him. Therefore, no matter how fast Achilles runs, he will never catch up with the tortoise.
"The slowest moving object can't catch up with the fastest passive object. Because the pursuer has to reach the point where the pursued should start first, the pursued has gone a long way. So the pursued person is always in front of the pursuer. "
── Aristotle
As Plato described it, Zhi Nuo said such a paradox was a little joke. First of all, parmenides fabricated this paradox to laugh at Pythagoras'1>; 0.999..., 1-0.999...& gt0 ". Then, he used this paradox to laugh at his student Zhi Nuo's thought of "1=0.999" ... but 1-0.999...& gt0 ". Finally, Zhi Nuo used this paradox to ridicule parmenides's "1-0.999" ... = 0, or1-0.999 ... >; 0 "thought.
Fixed arrow paradox
The arrow is stationary.
Because the arrow has its definite position at all times, it is static, so it can't be moving.
Paradox of parade
First of all, suppose that on the playground, at an instant (a minimum time unit), queues B and C will move to the right and left by a distance unit, respectively, relative to audience A.
□ Audience a
■■■■■ Queue B ... moves to the right.
▲▲▲▲▲▲▲▲ Queue C ... moves to the left.
Two lines b and c begin to move. As shown in the figure below, they moved one distance unit to the right and one distance unit to the left relative to audience A, B and C, respectively.
□□□□
■■■■
▲▲▲▲
At this point, for B, C has moved by two distance units. That is, the queue can move one distance unit instantly (a minimum time unit) or one distance unit within half a minimum time unit, which leads to the contradiction that half a time unit is equal to one time unit. So the queue cannot be moved.
Can Zeno Paradox be Solved by Sum of Infinite Series?
Peng Zheye (Jing Tian Man)
There is an idea that this problem can be solved by summation of infinite series (dichotomy and Achilles chasing turtles).
Let's assume that the space distance traveled by an object after it finally reaches its destination is 1, and the time distance traveled is 1. First of all, let's assume that the object has no last midpoint, so the distance S that the object travels in space after passing through infinite midpoints is:
S = 1/2+ 1/2 2+... 1/2n =(2n- 1)/2n = 1- 1/2。
As we can see, where s is the space distance 1 that is infinitely close to reality. However, infinite proximity does not mean that the object has not finally arrived.
Now let's assume that there is a final midpoint.
Then there is
s= 1/2+ 1/2^2+ 1/2^2
s= 1/2+ 1/2^2+ 1/2^3+ 1/2^3
.............
s= 1/2+ 1/2^2+ 1/2^3+......... 1/2^n+ 1/2^n
=(2^n- 1)/2^n+ 1/2^n= 1
That is to say, the distance traveled by an object after passing the distance between the last midpoint and the end point is consistent with the distance actually traveled by the object.
From the above calculation, we can simply see that if an object reaches the end point, it has passed the last midpoint. If it doesn't pass the last midpoint, it can't reach the finish line.
Similarly, we can calculate the time it takes for an object to pass through an infinite midpoint. Assume that the actual arrival time is 1. If the object does not have the last midpoint to pass through, the time t required for the object to pass through the infinite midpoint is:
t= 1/2+ 1/2^2+...... 1/2^n=(2^n- 1)/2^n= 1- 1/2^n
It can be seen that t here is the time required for an infinitely close object to actually reach the end point, but infinitely close is not equal to.
If there is one last midpoint for something, it is.
t= 1/2+ 1/2^2+ 1/2^3+......... 1/2^n+ 1/2^n
=(2^n- 1)/2^n+ 1/2^n= 1
That is to say, it takes the same time for an object to walk the distance between the last midpoint and the end point as it actually arrives.
It can be clearly seen from the above calculation that if an object has the last midpoint to go, then the time it takes is the same as the actual arrival time. If an object has no final midpoint to go, the time it takes can only be the time it takes for an infinitely close object to actually reach the end point, not equal to.
So the result of the summation of infinite series is that if things can reach the end point, they must go through the last midpoint. But how do things pass through the last midpoint? There is no basis here. In other words, the paradox of dichotomy still exists. In other words, this method of summing infinite series deepens the logic of this paradox. The dichotomy paradox and Achilles' tortoise-chasing paradox are actually two manifestations of the same paradox. Dichotomy can't be solved, but Achilles' pursuit of turtles remains the same.