Proof By Contradiction Worksheet Page 2
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1
2
3(#4) The details:
We say this:
p
By way of contradiction [BWOC, is an acceptable abbreviation], suppose x =
2 is rational.
We have assumed that (
Q) is true.
p
Thus, we can write
2 in rational form:
p
p
2 =
; where the fraction is in lowest terms.
q
We see that the “lowest terms” part is very important.
We square both sides.
2
p
2
2
2 =
) 2q
= p
:
2
q
2
From our odd/even lemmas, we see that the left side is even, and so p
must be even.
From a previous proof, we showed that p must also be even. Let p = 2k
:
1
2
2
2
2q
= (2k
)
= 4k
1
1
2
2
q
= 2k
:
1
2
Now the right side is even, and so q
must also be even. It follows that q must also be even.
CONTRADICTION! ()() [This is its symbol. Two arrows pointing at each other!]
If p and q are both even, then p=q is not in lowest terms.
Thus, (
Q) is false, and Q is true!
p
We have that the
2 is irrational, because it cannot be rational.
p
(#5) We can extend this technique for
3; but then we end up in this position:
2
2
2
3q
= p
) 3 j p
:
We would need to prove that
2
3 j p
) 3 j p:
Without the Prime Factorization Theorem, we have the cases:
p = 3k
1
p = 3k
+ 1
2
p = 3k
+ 2:
3
Obviously, only the …rst one works.
(#6) Here is the short form of the proof:
p
BWOC, assume that
2 is rational and that it can be written in lowest terms as:
p
2
p
p
2
2
2 =
) 2 =
) 2q
= p
:
2
q
q
Thus, p must be even and we rewrite it as p = 2k
1
2
2
2
2
2
2q
= (2k
)
= 4k
) q
= 2k
:
1
1
1
Thus, q must also be even and this contradicts p=q being in lowest terms. ()()
p
Therefore,
2 must be irrational.
2
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