There are no positive integer solutions to the equation
Contradiction
For the sake of contradiction suppose there are integer solutions to
Notice that
is even
is even
Now we have the same equation as the beginning
We can keep on doing this and have an infinite descent case.
Popular Set
1
Show that is transitive
By we know that
It's not reflexive
By we have that or but not both.
So if
We have and , which is a contradiction.
It's not symmetric
Implies that
Proved
2
If and then
so by it's not .
which shows us that it's in .
This is because by for every we have
So we can multiply.
If
So
3
If and
Then
By for every we have
By we know if somethings in , then the negative of it, is not.
So we know these are valid
We're done
4
If and show that
By for every we have
So if
This holds true for
So
5
If and show that
By we know that if then we have
Tutorial W3
1
Relation on a set
Equivalence if it's reflexive, symmetric and transitive
sign
Weak partial order if it's reflexive, anti-symmetric and transitive
sign
Strong partial order if it's irreflexive, anti-symmetric and transitive
sign
Define the relation on as
Reflexive
right now we can just say it's a strong partial order as it's irreflexive
Symmetric
Anti-symmetric
Because
So and
it must mean that
Which is the case
Transitive
Which means that
Which shows transitivity
So this is a strong partial order.
2
is a weak partial order on
Supremums and infimums
Define a relation on as
1
Why is and equivalence relation
If we take two sets, and the infimums and supremums are the same, as in the lowest upper bound and greatest lower bound are the same, then we have the same set.
Example:
Working in
we have
Also
This is why without proof, we can say it's an equivalence relation
Proof:
Reflexive
and the
This is trivially true
Symmetric
and
and the
and the
This obviously is true
It's not anti-symmetric because the elements in-between may differ
Example:
Working in a natural space
The supremums and infimums are the same, but the sets are not.
Transitive
and then
and the
and the
Want to show
and the
If and then
The same argument applies for
2
Describe the set of equivalence classes
This is any set from the power set of real number which relates to the given equivalence set .
3
Is the set in the equivalence class of under ?
The equivalence class is saying any set from the power set with lower bound and upper bound .
This is included as and the upper and lower bounds are the same as those of
Tutorial W2
1
Can these be true at the same time?
First let's check if the negation of one is the other
See that the second statement is the negation of the first, so this cannot be the same.
2
a
Rewrite and as and
If we are not in we are in
b
Part is true if and don't bisect the universe
Functions Worksheet
Suppose
is an arbitrary set
How is related to
Is one a subset of the other
is the image
is the preimage of the image
because of non-surjective functions
Consider the example
See that the preimage of the image results in only non-negative numbers.
Functions Exercises
1
Find and
is anything that maps to
for
2
as
is the set of primes
find and
, the radius of the circle is
's lower bound is
The upper bound is
Since primes have only and as factors
It would be
3
by
Find
I notice that with we will get as an output. However there is no value for which
As: which violates that the space under a must be non-negative
Otherwise,
6
a
Injective
Not injective
Counterexample:
True
However the inputs are not the same
Surjective
See that in the codomain has no element from the domain mapping to it.