Theorems on Rationals

We present three theorems involving rational numbers.

nnth Power Theorem. Let n n be a positive integer. If a a is a positive integer such that a=rn a = r^n for some rational number r r, then r r must be an integer.

Proof: We apply the integer root theorem to the polynomial xna x^n - a . Since this polynomial has a rational root r r, this root must be an integer. _\square

Theorem 2. If the sum and product of two rational numbers are both integers, then the two rational numbers must be integers.

Proof: We have r1+r2=b r_1 + r_2 = b and r1r2=c r_1 \cdot r_2 = c, hence r1,r2 r_1, r_2 are rational roots of the polynomial x2bx+c x^2 - bx + c. By the integer root theorem, r1 r_1 and r2 r_2 are both integers. _\square

Theorem 3. A quadratic polynomial f(x)=ax2+bx+c f(x) = ax^2 + bx +c has integer coefficients. Then both of the roots are integers if and only if:

1. b24ac b^2 - 4ac is a perfect square,

2. a a divides both b b and c c.

Proof: By the Remainder Factor Theorem, if the roots are integers n n and m m, then f(x)=a(xn)(xm) f(x) = a (x-n)(x-m) . By comparing terms, we obtain b=a(n+m) b = -a(n+m) and c=anm c = anm. Hence, (2) is satisfied. Furthermore, b24ac=a2(n+m)24a×anm=a2(nm)2, b^2 - 4ac = a^2 (n+m)^2 - 4 a \times anm = a^2 (n-m)^2, so (1) is also satisfied.

Conversely, by the quadratic formula, the roots of f(x)f(x) are b±b24ac2a \frac {-b \pm \sqrt{b^2- 4ac} }{2a} . By condition (1), it follows that the roots are rational. By condition (2) and Vieta's formula, both the sum and product of the roots are integers. Hence by Theorem 2, the roots are integers. _\square

Note by Calvin Lin
5 years, 8 months ago

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what's &nbsp?

Daniel Lim - 5 years, 3 months ago

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Something like a non-breaking space.

Edit: Actually, it's  

Sean Ty - 5 years, 3 months ago

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It was meant to be a non-breaking space, to separate out indented paragraphs. I wanted the theorems to be displayed individually, instead of in a long chunk of text.

Calvin Lin Staff - 5 years, 3 months ago

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