You needn't translate this to symbols to do this problem, but let me do so
to clarify the reasoning. First, let be the statement ``it rains'',
be ``I will wear a coat'', and be ``I'll stay home''. Thus, the
negation of the statement is
When it rains, I won't wear a coat and I won't stay home.or perhaps you might say
I will go out in the rain without a coat.
Solution: This is of the form , which is negated as , so we get
This function is invertible or it is continuous.I suppose if you wanted to write this as an implication, you could write
If this function has no inverse, then it is continuous.which is equivalent.
Solution: Here we have a statement of the form ; the negation of such a statement is of the form . Thus, we get
There is a triangle where the sum of the measure of its angles is at least .
This statement is nearly already in symbolic form. The negation is
You may recall from calculus that the original statement is the definition of ``The function is continuous at the point '', so the latter is `` is not continous at ''.
There is an implied quantifier on the in the statement, that is, it
really should say
Solution: The negation is clearly
Not every natural number has a unique additive inverse.or
There is a natural number which does not have a unique additive inverse.
But we can elaborate further. If a number does not have a unique inverse, either it doesn't have an inverse at all, or it has more than one.2 So, we have
There is a natural number which has no additive inverse, or there is a natural number with more than one additive inverse.
Solution: This is false. Consider and . Then , which is positive, but , which is not negative.
Since we are insisting that everything be proven with reference to axioms, we probably should work a little harder to fully justify all of our statements. (On an exam, I probably would give nearly full credit for the above counterexample alone. But this isn't an exam, so I should be complete.)
First we know that and exist, as consequences of axioms V.8 and V.9, and the fact that follows immediately from axiom V.8. But, we have not shown that is positive, nor have we shown that .
One way to see that is to use axiom VI.1 ( ), but perhaps this is a bit of a cheat, because we didn't define ``positive'' in that way.
So let's prove by contradiction.
Suppose not. Then either or . But contradicts axiom V.12.
If , then (see the Lemma below). Then applying axiom V.12 (with , , and ), we get . But this says , contradicting the earlier statement. Hence, , that is, is a positive number.
To see that , let's suppose that there is number which is the additive inverse of . Then by axiom V.10, and by axiom V.8 (and V.5). Thus, , so is its own additive inverse.
Finally, we need the following.
Lemma. For all
Proof. Suppose . Then by axiom V.16, , and applying axiom V.8 to the right and V.10 to the left, we get .
Similarly, if , we have , so .
Whew! More than I bargained for here.
Solution: True. Since we need merely demonstrate existence of such and , let and . Then is positive (we just did this), and is negative (again, see the previous part.)
Solution: This is true. We must show that given a specific integer , we can find an integer that meets our needs. Let's break the problem up into three possibilities: , , and .
Case I: . Let . First, notice that , since and so by axiom V.16, . By transitivity of (axiom V.14), we have . Thus, again by V.16, . Thus (by transitivity again) , so is positive.
We also need show that is negative. Note that (distributive law, commutivity), and so . (using associativity, additive identity). But is negative, from before.
Case II: . Now let . Then , and is positive. To see that , we have . Using the fact that and , together with transitivity and axioms V.14 and V.16, we get .
Case III: . Easiest of all. Let , so , which is positive, and , negative.
Solution: This is false. This says that we can find this , and write it down somewhere. Now for this and any possible , must be positive and negative. What about the choice ? It is supposed to be positive, but , which isn't positive.
Another way to see this is to write it in symbols. This statement is
But this follows from the previous part. Since the negation is true, the statement can't be.
Solution: Axiom K1 says
For every person, either the person is male or female, but that person cannot be both male and female.This is better said as
Every person is either male or female, but not both.
For every person , there is a unique person and a unique person so that and are parents of , is male, and is female.This is a very stilted way to say
Everybody has exactly one father and exactly one mother.provided that we know that ``father'' means ``male parent'' and ``mother'' means ``female parent''.
is a grandfather ofsince is a parent of and is 's father.
Everybody has a grandfather.
Solution: Let be an arbitrary person. Then by axiom , has a father, that is, there is some for which and . Now apply axiom again to this new to get 's father, who we shall denote . Since 's father's father is his/her grandfather, we have . (Of course, there is also the maternal grandfater). We have shown that for this , . Since was arbitrary, we apply UG to conclude that .
We want to establish that
, where is
the statement `` is divisible by . Using induction as
suggested, we first check :
Is divisible by ? Since , yes it is.
Now we show that , that is, we want to show is divisible by if we know that is.
If is divisible by 3, then there is an so that .
Consequently, . So, we have