Äquivalenzrelationen
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
Short
Video
\(\LaTeX\)
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Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
Sei sim eine Äquivalenzrelation auf der Menge mathbbN... derart dass a sim a+ wedge a sim a+ für alle a in mathbbN gilt. Gilt sim ? Wie viele Elemente hat der Quotient mathbbN/sim?
Solution:
Wir zeigen zuerst dass die Äquivalenzen asim a+ und a sim a+ implizieren dass a sim a + für alle a in mathbbN. In der Tat ist: a sim a + sim a + sim a + sim a + sim a + wobei wir die Symmetrie benutzt haben um a + sim a für alle a in mathbbN zu benutzen. Aus mehrfacher Anwung der Transitivität folgt nun a sim a + für alle a in mathbbN. Als nächstes folgern wir dass a sim b für alle a b in mathbbN. Wegen der Symmetrie ist es genug zu zeigen dass a sim b für alle a b in mathbbN mit a leq b. Für solche a und b gilt: a sim a + sim ... sim b - sim b Wieder mehrfache Nutzung der Transitivität gibt a sim b für alle a b in mathbbN. Also sind alle Elemente aus mathbbN bezüglich dieser Äquivalenzrelation zueinander äquivalent das heisst die Menge der Äquivalenzklassen mathbbN/sim besteht nur aus einem Element.
Sei sim eine Äquivalenzrelation auf der Menge mathbbN... derart dass a sim a+ wedge a sim a+ für alle a in mathbbN gilt. Gilt sim ? Wie viele Elemente hat der Quotient mathbbN/sim?
Solution:
Wir zeigen zuerst dass die Äquivalenzen asim a+ und a sim a+ implizieren dass a sim a + für alle a in mathbbN. In der Tat ist: a sim a + sim a + sim a + sim a + sim a + wobei wir die Symmetrie benutzt haben um a + sim a für alle a in mathbbN zu benutzen. Aus mehrfacher Anwung der Transitivität folgt nun a sim a + für alle a in mathbbN. Als nächstes folgern wir dass a sim b für alle a b in mathbbN. Wegen der Symmetrie ist es genug zu zeigen dass a sim b für alle a b in mathbbN mit a leq b. Für solche a und b gilt: a sim a + sim ... sim b - sim b Wieder mehrfache Nutzung der Transitivität gibt a sim b für alle a b in mathbbN. Also sind alle Elemente aus mathbbN bezüglich dieser Äquivalenzrelation zueinander äquivalent das heisst die Menge der Äquivalenzklassen mathbbN/sim besteht nur aus einem Element.
Meta Information
Exercise:
Sei sim eine Äquivalenzrelation auf der Menge mathbbN... derart dass a sim a+ wedge a sim a+ für alle a in mathbbN gilt. Gilt sim ? Wie viele Elemente hat der Quotient mathbbN/sim?
Solution:
Wir zeigen zuerst dass die Äquivalenzen asim a+ und a sim a+ implizieren dass a sim a + für alle a in mathbbN. In der Tat ist: a sim a + sim a + sim a + sim a + sim a + wobei wir die Symmetrie benutzt haben um a + sim a für alle a in mathbbN zu benutzen. Aus mehrfacher Anwung der Transitivität folgt nun a sim a + für alle a in mathbbN. Als nächstes folgern wir dass a sim b für alle a b in mathbbN. Wegen der Symmetrie ist es genug zu zeigen dass a sim b für alle a b in mathbbN mit a leq b. Für solche a und b gilt: a sim a + sim ... sim b - sim b Wieder mehrfache Nutzung der Transitivität gibt a sim b für alle a b in mathbbN. Also sind alle Elemente aus mathbbN bezüglich dieser Äquivalenzrelation zueinander äquivalent das heisst die Menge der Äquivalenzklassen mathbbN/sim besteht nur aus einem Element.
Sei sim eine Äquivalenzrelation auf der Menge mathbbN... derart dass a sim a+ wedge a sim a+ für alle a in mathbbN gilt. Gilt sim ? Wie viele Elemente hat der Quotient mathbbN/sim?
Solution:
Wir zeigen zuerst dass die Äquivalenzen asim a+ und a sim a+ implizieren dass a sim a + für alle a in mathbbN. In der Tat ist: a sim a + sim a + sim a + sim a + sim a + wobei wir die Symmetrie benutzt haben um a + sim a für alle a in mathbbN zu benutzen. Aus mehrfacher Anwung der Transitivität folgt nun a sim a + für alle a in mathbbN. Als nächstes folgern wir dass a sim b für alle a b in mathbbN. Wegen der Symmetrie ist es genug zu zeigen dass a sim b für alle a b in mathbbN mit a leq b. Für solche a und b gilt: a sim a + sim ... sim b - sim b Wieder mehrfache Nutzung der Transitivität gibt a sim b für alle a b in mathbbN. Also sind alle Elemente aus mathbbN bezüglich dieser Äquivalenzrelation zueinander äquivalent das heisst die Menge der Äquivalenzklassen mathbbN/sim besteht nur aus einem Element.
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