Compositions of functions
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:
Let f:X rightarrow Y g:Y rightarrow Z be maps. The following properties are to be shown: abcliste abc If fg are injective then g circ f is also injective. abc If fg are surjective then g circ f is also surjective. abc If fg are bijective then g circ f is also bijective. abcliste
Solution:
Proof. abcliste abc We need to prove that forall x_ x_ in X the following holds: if gcirc fx_gcirc fx_ then x_x_. So let x_x_ in X be s.t. gcirc fx_gcirc fx_. This is equal to gfx_gfx_. Since g is injective it follows that fx_fx_. Since g is injective as well we can conclude that x_x_. abc We need to show that forall z in Z exists x in X: gcirc fxz. Indeed let z in Z. Since g is surjective exists y in Y: gyz. As f is also surjective it follows that exists x in X: fxy. This allows the conclusion: gcirc fx gfxgyz. abc Follows from a+b because glqq bijective injective+surjectivegrqq abcliste
Let f:X rightarrow Y g:Y rightarrow Z be maps. The following properties are to be shown: abcliste abc If fg are injective then g circ f is also injective. abc If fg are surjective then g circ f is also surjective. abc If fg are bijective then g circ f is also bijective. abcliste
Solution:
Proof. abcliste abc We need to prove that forall x_ x_ in X the following holds: if gcirc fx_gcirc fx_ then x_x_. So let x_x_ in X be s.t. gcirc fx_gcirc fx_. This is equal to gfx_gfx_. Since g is injective it follows that fx_fx_. Since g is injective as well we can conclude that x_x_. abc We need to show that forall z in Z exists x in X: gcirc fxz. Indeed let z in Z. Since g is surjective exists y in Y: gyz. As f is also surjective it follows that exists x in X: fxy. This allows the conclusion: gcirc fx gfxgyz. abc Follows from a+b because glqq bijective injective+surjectivegrqq abcliste
Meta Information
Exercise:
Let f:X rightarrow Y g:Y rightarrow Z be maps. The following properties are to be shown: abcliste abc If fg are injective then g circ f is also injective. abc If fg are surjective then g circ f is also surjective. abc If fg are bijective then g circ f is also bijective. abcliste
Solution:
Proof. abcliste abc We need to prove that forall x_ x_ in X the following holds: if gcirc fx_gcirc fx_ then x_x_. So let x_x_ in X be s.t. gcirc fx_gcirc fx_. This is equal to gfx_gfx_. Since g is injective it follows that fx_fx_. Since g is injective as well we can conclude that x_x_. abc We need to show that forall z in Z exists x in X: gcirc fxz. Indeed let z in Z. Since g is surjective exists y in Y: gyz. As f is also surjective it follows that exists x in X: fxy. This allows the conclusion: gcirc fx gfxgyz. abc Follows from a+b because glqq bijective injective+surjectivegrqq abcliste
Let f:X rightarrow Y g:Y rightarrow Z be maps. The following properties are to be shown: abcliste abc If fg are injective then g circ f is also injective. abc If fg are surjective then g circ f is also surjective. abc If fg are bijective then g circ f is also bijective. abcliste
Solution:
Proof. abcliste abc We need to prove that forall x_ x_ in X the following holds: if gcirc fx_gcirc fx_ then x_x_. So let x_x_ in X be s.t. gcirc fx_gcirc fx_. This is equal to gfx_gfx_. Since g is injective it follows that fx_fx_. Since g is injective as well we can conclude that x_x_. abc We need to show that forall z in Z exists x in X: gcirc fxz. Indeed let z in Z. Since g is surjective exists y in Y: gyz. As f is also surjective it follows that exists x in X: fxy. This allows the conclusion: gcirc fx gfxgyz. abc Follows from a+b because glqq bijective injective+surjectivegrqq abcliste
Contained in these collections
| Title | Matched on |
|---|---|
| Linearity Fibonacci sequences | tags |
| Row equivalence and solutions | tags |
| Group characteristics | tags |
| Russel paradox | tags |
| Algorithm to find a basis | tags |