We covered from Proposition 1.1.4 about the inequality relation — up to but not including Proposition 1.4.3.
I will make a decision about tutorial clashes on or about Monday.
Exercises
Q. 1,2 and 4 – 7 from http://euclid.ucc.ie/pages/staff/wills/teaching/ms2001/exercise1.pdf
Q. 2 – 3 from Problems
J.P. McCarthy: Math Page 
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October 5, 2011 at 11:06 pm
J.P. McCarthy
Question 5
is both even and odd. That is for all 
we have that
and 
. Oddness implies that 
as 
as 
. However 
is the only number equal to its own negative so that 
.
. By evenness and then oddness 
and 
. Combining these gives 
. Once again the only number that satisfies this is 
so 
for all 
. Now by evenness 
so that the only function which is odd and even is the function 
.
Suppose that
Now suppose that
Question 6
is even we must show that 
. Hence examine 
as 
is odd. However we can show that 
so therefore 
; that is 
is even.
To show that
Even and odd functions have a class of symmetry about
. If we kinda break this we can come up with some function 
that is neither even nor odd but has the property that 
is even. All constant functions are even so if we could find a non-symettric function that is equal to 
then we’d be done as 
. Choose
— this function has the desired property.