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Math 2534____In Class Lecture-Worksheet for Monday (Sept 20)
Answers:
Prove the following:
1) Let x and y be Real Numbers.
a) Use a direct proof to prove that if x = y, then x2 = y2 .
b) Redo the proof in part a) using proof by contradiction.
Solution: a) For x and y , real numbers, prove that if x = y then x2 = y2
Proof: Since x = y, we have that xx = yx and therefore
xx = y(y) to give us x2 = y2 . There fore if x = y then x2 = y2
b) Proof by contradiction: Given that x = y and assume
that x2 ≠ y2 . If x2 ≠ y2 Then x2 - y2 ≠ 0 and
(x - y)(x + y) ≠ 0 Therefore (x – y) ≠ 0 and/or (x + y) ≠ 0.
The only way this can be true is for x ≠ y which is a
contradiction to our hypothesis. Therefore the original
conjecture must be true.
2) Use the fifth postulate of Euclid and proof by contradiction to prove the
following:
Let u, v, and w be distinct lines in a plane. If u is parallel to v and v is
parallel to w, then u is parallel to w.
Fifth Postulate: Given a line m and a point p not on m (in a plane), there is
exactly one line through p and parallel to m.
Solution: Let u, v, and w be distinct lines in a plane. If u is parallel to v
and v is parallel to w, then u is parallel to w.
Proof by contradiction: Assume u is not parallel to w then u and
w intersect in some point p. Since by the Fifth Postulate, only one line
can pass through p and be parallel to v. Therefore u and v must be the
same line. But this contradicts that u and w are distinct lines. The
original statement is true and u and w are parallel.
3) Prove by contradiction: if x is a real number and x3 + 4x = 0, then x = 0
Solutions: If x is a real number and x3 + 4x = 0, then x = 0
Proof by contradiction: we will assume that x ≠ 0, If we consider
x3 + 4x , we have x (x2 + 4) and since x ≠ 0, then x (x2 + 4) ≠ 0
which contradicts our hypothesis that x3 + 4x = 0. There x = 0.
4) Use proof by contradiction to prove that for all integers m, 7m + 4 is not
divisible by 7.
Solutions: 7m + 4 is not divisible by 7 for any integer m.
Proof by contradiction: Assume that 7m + 4 is divisible by 7.
Then by the definition of divisibility we have that 7m + 4 = 7q for some
integer q. After some calculations we have that 7m -7q = -4 and
7(m-q) =- 4 giving us that m-q = -4/7. This is not possible since the
difference of two integers is always an integer. Therefore 7m+4 is not
divisible by 7
5) Use the method of contradiction to prove that 3 is irrational.
Proof by contradiction: Assume that 3 is rational. Then by
definition of rational 3 = a/b for some integers a, b ≠ 0. We will assume
that a/b is in lowest reduced terms. We square both sides to get 3 =
a2
.
b2
Now we have that 3b2=a2 and a2 is divisible by 3 by definition. Since a2
is divisible by 3, we have that a is divisible by 3 since 3 is a prime and a2
factors into a*a. This gives a = 3q for some integer q.
We now have that 3b2=a2 = 9q2 and b2 = 3q2. So we have that b2 is also
divisible by 3. By the same argument for “a”, we have that b is divisible
by 3 giving us that b = 3d for some integer d. This means that 3 = a/b
= 3q/3d is not in lowest terms. This is a contradiction to our assumption
so 3 must be irrational.