The Euclidean Algorithm

Home , Long division

1. The Euclidean algorithm. The euclidean algorithm is used to ﬁnd GCD(a, b), the greatest common divisor of two positive integers a, b ∈ N, and can be used to ﬁnd m, n ∈ Z so that:

GCD(a, b) = ma + nb.

In particular, if p is a prime, GCD(a, p) = 1, so writing 1 = ma + np we −1 ﬁnd that in the ﬁeld Zp (of residues mod(p)) we have: (¯a) =m ¯ . ¯ Example. Find the multiplicative inverse of 11 in Z37. The euclidean algorithm gives:

37 = 3 × 11 + 4, 11 = 2 × 4 + 3, 4 = 3 + 1.

Thus:

1 = 4 − 3 = 37 − 3 × 11 − [11 − 2 × (37 − 3 × 11)] = 3 × 37 − 10 × 11,

¯ −1 ¯ ¯ or m = −10, n = 3. Thus (11) = −10 = 27 in Z37. ¯ −1 Problem 1. Find (4) in the ﬁeld Z31. 2. Long division. Suppose x ∈ [0, 1) has an eventually periodic decimal representation: x = 0.m1m2 . . . mN d1d2 . . . dT . Then:

N+T N 10 x − m1 . . . mN d1 . . . dT = 10 x − m1 . . . mN = 0.d1 . . . dT .

Hence:

N T 10 (10 − 1)x = m1 . . . mN d1 . . . dT − m1 . . . mN := m ∈ N, or: m d . . . d x = (x = 1 T if the expansion is periodic). 10N (10T − 1) 10T − 1

In other words: any real number (in [0, 1)) with an eventually periodic decimal expansion is rational, and can be written in the form m/q, where q is a diﬀerence of powers of 10. (Or, considering base 2 expansions, with q a diﬀerence of powers of two.)

1 And yet any rational number has an eventually periodic decimal expansion! This can be established by considering long division. Consider x = p/q with p, q ∈ N, p < q. We have (interpreting the long division algorithm):

10p = d1q + r1, 10r1 = d2q + r2, 10r2 = d3q + r3,...

p and this gives the decimal expansion: q = 0.d1d2d3 .... The sequence of remainders takes values in the ﬁnite set {0, 1, . . . , q − 1}, so eventually a remainder has to be repeated, and after that the whole sequence of digits in the decimal expansion repeats. Thus the expansion is eventually periodic, with period T ≤ q − 1 (if a remainder is zero, this means the expansion is ﬁnite, so either eventually 0’s or eventually 9’s, depending on which one you choose.) 2 Example. (seen in class) 7 = 0.285714. This is gotten from long division: 20 = 2×7+6, 60 = 8×7+4, 40 = 5×7+5, 50 = 7×7+1, 10 = 1×7+3, 30 = 4×7+2.

This implies: 2 285714 = , hence 7 divides 106 − 1 = 999, 999. 7 106 − 1

2 2 We can get the binary expansion easily if we observe that 7 = 23−1 : it is periodic with period 3. Since 2 written in base 2 is 10, we have: 2 ( ) = 0.010. 7 base 2 Problem 2. (i) Use the eventually periodic expansion of 5/7 to write it in the form m/q, where q is a difference of powers of 10. (ii) Write down the binary expansion of 5/7. √ Problem√ 3. Find the multiplicative inverse of x = −3 + 5 7 in the field Q( 7). Problem 4. Consider two arbitrary polynomials of degree 2 and integer coefficients:

2 2 p(x) = a0 + a1x + a2x , q(x) = b0 + b1x + b2x , aj, bj ∈ Z. Show that if their product p(x)q(x) (of degree 4) is the zero polynomial (that is, has all its coeﬃcients 0), then at least one of p(x), q(x) is the zero polynomial. (Hint: Show that a0b0 = 0 and a2b2 = 0. This gives four cases to consider).

2 This problem shows that the fact that the ring Z[x] is an integral domain (i.e., has no zero divisors) has a non-trivial proof.

Recall that a commutative ring R is an integral domain if: (∀x, x¯ ∈ R)xx¯ = 0 → (x = 0 ∨ x¯ = 0). A commutative ring is a ﬁeld if every non-zero element has a multiplicative inverse: (∀x ∈ R)x 6= 0 → (∃x0 ∈ R)xx0 = 1. Problem 5. Show that any ﬁeld is an integral domain.