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

### Like this:

Like Loading...

*Related*

## 1 comment

Comments feed for this article

October 5, 2011 at 11:06 pm

J.P. McCarthyQuestion 5

Suppose that 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 .

Now suppose 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 .

Question 6

To show that is even we must show that . Hence examine as is odd. However we can show that so therefore ; that is is even.

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.