Rational and Irrational Numbers
Original Notes adopted from October 23, 2001 (Week 7)
© P. Rosenthal , MAT246Y1, University of Toronto, Department of Mathematics typed by A. Ku Ong
Lemma: If x2 is even then x is even.
Theorem: ¥ 2 is irrational
We can’t have ¥ 2 = m / n or 2 = (m / n) 2 for m, n integers
Proof: If 2 = (m/n) 2 , we can divide out all common factors of m & n ,
getting m0/n0 in lowest terms.
(m0/n0) 2 =2 ⇒ m02 = 2n02 ∴ m02 is even ⇒ m0 is even (from Lemma)
∴ m0 = 2k, some k. (m0 is even)
(m0/n0) 2 =2 ⇒ m02 = 2n02 , substitute m0 = 2k, some k (from above)
(2k) 2 = 2n02
4k2 = 2n02
∴n02 is even ⇒ n0 is even (from Lemma)
2k2 = n02
Now since 2|n0, 2|m0 ⇒ m0/n0 is not in lowest terms so ¥ 2 is NOT
rational.
Theorem: ¥ 3 is irrational
Proof: Assume rational ¥ 3 = m/n means m/n are in lowest terms ( We will
show there is a contradiction)
3 = m2/n2 (square both sides)
3n2 = m2
(Note since 3 is prime theorem states p|ab ⇒ p|a or p|b)
Note : 3 | m * m ⇒ 3|m or 3 |m.
3 |m. ⇒ so substitute m = 3k, some k
3n2 = (3k) 2
3n2 = 9k 2
n2 = 3k 2
Note : 3|n*n ⇒ 3|n or 3|n.
Both m and n are divisible by 3, so NOT in lowest terms Contradiction.
Theorem: If p is prime then ¥ p is irrational
Proof: Assume rational ¥ p = m/n means m/n are in lowest terms ( We will
show there is a contradiction)
p = m2/n2 (square both sides)
pn2 = m2
(Note since p is prime theorem states p|ab ⇒ p|a or p|b)
Note : p | m * m ⇒ p|m or p |m.
p |m. ⇒ so substitute m = pk, some k
pn2 = (pk) 2
pn2 = p2k 2
n2 = pk 2
Note : p|n*n ⇒ p|n or p|n.
Both m and n are divisible by p, so NOT in lowest terms Contradiction.
Notice ¥ 6 is irrational
Proof: Assume rational ¥ 6 = m/n means m/n are in lowest terms ( We will
show there is a contradiction)
6 = m2/n2
6n2 = m2
(square both sides)
Notice 2|m, 3|m ⇒ 6|m ⇒ m is not in lowest terms.
Substitute m = 6k, some k ⇒ 6n2 = (6k) 2 ⇒ 6n2 = 36k 2 ⇒ n2 = 6k 2
Notice: 6|n ⇒ n is not in lowest terms. ∴m/n is not in lowest
terms, contradiction.
Note: An integer of the form k2 for k an integer is a “perfect square”
Question: If ¥ n is rational, must n be a perfect square?
Ie. If ¥ n is rational, must it be an integer?
An Integer expressed as primes
Any integer > 1 can be written uniquely (except for order) in the form p1α1 ,
p2α2 …. pLαi
Where the {pi} are distance primes and {αi} are natural numbers.
Perfect Squares
A number is a perfect square exactly when all the powers of its primes are
even.
If N is a perfect square (N ≠ 1)
N = (p1α1 , p2α2 …. pLαi )2
= p12α1 , p22α2 …. pL2αi
Theorem: If N integer > 1, then ¥N is irrational unless N is a perfect
square.
Proof: Suppose N = (m/n)2 (Recall all primes occur evenly)
n2N=m2
Factor each of n, N, m into primes:
n = p1α1 …… p uα u
N = q1β1 …… q r β v
m = r1Μ1 ……..r tΜ t
(p1α1 …… p uα u) 2 (q1β1 …… q r β v ) = (r1Μ1 ……..r tΜ t)2
Μ1
Μt
…….. r t2Μ
(*) p12αα1 …… p u2αα u (q1β1 …… q r β v ) = r12Μ
To show N is a perfect square, we must show β1, β2 …. βn are all even.
To show βi is even: qi is some rj (unique factorization).
Then the occurrence of q1 on left side of (*) is to the
power 2Μj
Case 1: No pi is qi. The only occurrence of qi is qi, so β even (= 2Μj)
Case 2: Some pi = qi. Then on the left we have: p12α1qiβi = rj2αj
rj2αirjβI = rj2Μj
2αi + β i = 2 Μ j ∴ β is even.
square.
Etc ⇒ β L even for all L ∴ N is a perfect
Is t 3¥4 irrational?
3
¥4 = m/n , 4 = m3/n3 Notice: m = (p1α1 …… p
4n3 = m3 ⇒ 22n3 = m3 ⇒ impossible.
k
αk 3
) = p13α1 …… p
k
3α k