Lesson 2: Which is Greater?

Property

To eliminate nested radicals like $\sqrt{a + \sqrt{b}}$, square both sides repeatedly while preserving inequality direction. If $x > 0$ and $y > 0$, then $x > y$ if and only if $x^2 > y^2$.

Examples

Property

To compare two exponential numbers like $a^m$ and $b^n$, we can simplify their exponents by taking a common root. Let $k = \text{gcf}(m, n)$. The comparison between $a^m$ and $b^n$ is equivalent to comparing their $k$-th roots.

This reduces the exponents to smaller, more manageable numbers.

Examples

Example 1
Which is greater: 3^400 or 5^240?
The exponents are 400 and 240. The greatest common factor is gcf(400, 240) = 80. We compare (3^400)^(1/80) and (5^240)^(1/80). This simplifies to comparing 3^5 and 5^3, which is 243 vs 125. Since 243 > 125, we have 3^400 > 5^240.

Example 2
Which is greater: $5^{48}$ or $10^{36}$?
The exponents are $48$ and $36$. The greatest common factor is $\text{gcf}(48, 36) = 12$. We compare $5^{\frac{48}{12}}$ and $10^{\frac{36}{12}}$. This simplifies to comparing $5^4$ and $10^3$, which is $625$ vs $1000$. Since $625 < 1000$, we have $5^{48} < 10^{36}$.

Property

When comparing numbers very close to 1, express each as $1 - \epsilon$ where $\epsilon$ is a small positive quantity. If $a = 1 - \epsilon_1$ and $b = 1 - \epsilon_2$, then $a > b$ if and only if $\epsilon_1 < \epsilon_2$.

Examples

Property

For any positive numbers $a$ and $b$, the direction of an inequality is reversed when we take the reciprocal of both sides. If $a > b$, then $\frac{1}{a} < \frac{1}{b}$.

Examples

Example 1
To compare $A = \frac{100}{101}$ and $B = \frac{101}{102}$, we compare their reciprocals.
$\frac{1}{A} = \frac{101}{100} = 1 + \frac{1}{100}$
$\frac{1}{B} = \frac{102}{101} = 1 + \frac{1}{101}$
Since $\frac{1}{100} > \frac{1}{101}$, we have $\frac{1}{A} > \frac{1}{B}$, which implies $A < B$.
Example 2
Compare $a = \frac{\sqrt{5}}{\sqrt{5}+1}$ and $b = \frac{\sqrt{6}}{\sqrt{6}+1}$.
$\frac{1}{a} = \frac{\sqrt{5}+1}{\sqrt{5}} = 1 + \frac{1}{\sqrt{5}}$
$\frac{1}{b} = \frac{\sqrt{6}+1}{\sqrt{6}} = 1 + \frac{1}{\sqrt{6}}$
Because $\sqrt{5} < \sqrt{6}$, we know $\frac{1}{\sqrt{5}} > \frac{1}{\sqrt{6}}$.
Therefore, $\frac{1}{a} > \frac{1}{b}$, so $a < b$.

Explanation

When comparing two fractions that are very close to 1, it can be easier to compare their reciprocals instead. By taking the reciprocal of each number, you can often express them in the form $1 + \epsilon$, where $\epsilon$ is a small fraction. Comparing the sizes of these small fractions allows you to determine the larger reciprocal. Remember that if the reciprocal of one number is larger, the original number itself is smaller.