The Fabric of Fractions
A rational number is any number that can be expressed as a fraction \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \neq 0 \). The set of rational numbers is denoted by \( \mathbb{Q} \).
One of the most fascinating properties of \( \mathbb{Q} \) is that it is dense in \( \mathbb{R} \). This means that between any two real numbers, no matter how close, there is a rational number.
Visualizing Density
Drag the slider to increase the "resolution" of the rational numbers. We are plotting all fractions \( \frac{p}{q} \) where \( q \le \text{Max Denominator} \). Notice how the gaps get smaller and smaller, yet they never truly disappear (that's where the irrationals live!).
Rational vs. Irrational: The Cycle of Remainders
Rational numbers always have terminating or repeating decimal expansions. This is because when performing long division by \( q \), there are only \( q \) possible remainders. Eventually, a remainder must repeat!
Remainder Cycle Visualization
Each dot represents a possible remainder (0 to q-1).
Lines show the path of the long division.
Practice Problems
True/False
1. If \( \sqrt{17} \) satisfies \( x^2 - 17 = 0 \), then the only possible rational solutions are \( \pm1, \pm17 \).
Show Solution
TRUE - For \( x^2 - 17 = 0 \) (monic polynomial), rational solutions must be integers dividing -17.
2. Every irrational-looking expression like \( \sqrt{4+2\sqrt{3}} \) is actually irrational.
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FALSE - Some expressions simplify to rationals. Example: \( \sqrt{4+2\sqrt{3}} = 1 + \sqrt{3} \) is irrational, but \( \sqrt{4+2\sqrt{3}} - \sqrt{3} = 1 \) is rational.
3. For \( 3x^2 - 7 = 0 \), if there's a rational solution \( \frac{c}{d} \) in lowest terms, then \( c \) must divide -7 and \( d \) must divide 3.
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TRUE - This is the Rational Zeros Theorem.
4. If a number satisfies a polynomial equation with integer coefficients, it must be rational.
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FALSE - Such numbers are called algebraic, but can be irrational (e.g., \( \sqrt{2} \)).
Fill in the Blank
5. Prove \( \sqrt{3} \) is irrational.
\( \sqrt{3} \) is a solution of \( x^2 - 3 = 0 \).
By Rational Zeros Theorem, any rational solution must be an integer that divides _____.
Possible solutions: _____.
Testing: \( (\pm 1)^2 - 3 = \) _____, \( (\pm 3)^2 - 3 = \) _____.
Since none work, \( \sqrt{3} \) cannot be _____.
Show Solution
Divides: -3
Possible: \( \pm 1, \pm 3 \)
Values: -2, 6
Cannot be: rational
6. Prove \( \sqrt[3]{5} \) is irrational.
Satisfies \( x^3 - 5 = 0 \).
Possible rational solutions divide _____. These are: _____.
Test: \( 1^3 - 5 = \) _____, \( (-1)^3 - 5 = \) _____.
Therefore, \( \sqrt[3]{5} \) is _____.
Show Solution
Divides: -5
Candidates: \( \pm 1, \pm 5 \)
Values: -4, -6
Is: irrational
7. For \( 49x^4 - 56x^2 + 4 = 0 \), list all possible rational solutions.
\( c \) must divide \( c_0 = \) _____, so \( c \in \{ \)_____ \( \} \)
\( d \) must divide \( c_4 = \) _____, so \( d \in \{ \)_____ \( \} \)
All candidates: _____
Show Solution
\( c_0 = 4 \), \( c \in \{ \pm 1, \pm 2, \pm 4 \} \)
\( c_4 = 49 \), \( d \in \{ \pm 1, \pm 7, \pm 49 \} \)
Candidates: \( \pm 1, \pm 2, \pm 4, \pm 1/7, \pm 2/7, \pm 4/7, \pm 1/49, \pm 2/49, \pm 4/49 \)
Full Problems
8. Prove \( \sqrt{5} \) is not a rational number.
Show Solution
\( \sqrt{5} \) satisfies \( x^2 - 5 = 0 \). Candidates: \( \pm 1, \pm 5 \).
None satisfy the equation (e.g., \( 1^2-5=-4 \)). Thus irrational.
9. Prove \( \sqrt[3]{2} \) is not a rational number.
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\( x^3 - 2 = 0 \). Candidates: \( \pm 1, \pm 2 \).
None work. Irrational.
10. Show that \( \sqrt{2} + \sqrt{5} \) is irrational.
Show Solution
Let \( a = \sqrt{2} + \sqrt{5} \). \( a^2 = 7 + 2\sqrt{10} \). \( (a^2-7)^2 = 40 \).
\( a^4 - 14a^2 + 9 = 0 \). Candidates: \( \pm 1, \pm 3, \pm 9 \).
None work. Irrational.
11. Consider \( 2x^3 - 5x + 3 = 0 \). List candidates and find solutions.
Show Solution
Candidates: \( \pm 1, \pm 3, \pm 1/2, \pm 3/2 \).
\( x=1 \) is a solution.
12. Prove \( \sqrt{12} \) is irrational.
Show Solution
\( \sqrt{12} = 2\sqrt{3} \). If rational, \( \sqrt{3} \) would be rational. Contradiction.