The Irrational Number Search
Explore the concept of irrational numbers! Your goal is to see if you can land a fraction point exactly on the target value, the square root of 2.
Help & Instructions
â–¼- *Target Value:* The red dot represents the exact location of the square root of 2 on the number line.
- *Adjust Fraction:* Enter different numerators and denominators to create a fraction.
- *Find a Match:* The green dot will move to the location of your fraction. Try to make the green dot land exactly on the red dot.
- *Observe:* No matter what whole numbers you choose for the fraction, you will not be able to land on the red dot.
- Understand the difference between rational and irrational numbers.
- Visually grasp why an irrational number cannot be represented as a simple fraction.
- Recognize that the decimal expansion of irrational numbers is non-repeating and non-terminating.
Target Value: √2 ≈ 1.4142...
Current Fraction: 0/1
A *rational number* is any number that can be expressed as a fraction $p/q$, where $p$ and $q$ are integers and $q$ is not zero. An *irrational number* is a real number that cannot be expressed as a simple fraction. This is because their decimal representation is non-repeating and non-terminating.
The Mathematics Behind the Concept
The **square root of 2**, or $\sqrt{2}$, is a famous example of an irrational number. The ancient Greeks, particularly the Pythagoreans, discovered that the diagonal of a square with a side length of 1 unit cannot be measured by any simple fraction. This was a profound discovery that showed the number line contains more than just rational numbers.
A simple proof shows that $\sqrt{2}$ is irrational. Assume $\sqrt{2}$ is rational, meaning it can be written as a fraction $p/q$ in its simplest form. Squaring both sides gives $2 = p^2/q^2$, or $2q^2 = p^2$. This means $p^2$ is an even number, which implies $p$ must also be even. We can write $p=2k$ for some integer $k$. Substituting this back gives $2q^2 = (2k)^2 = 4k^2$, which simplifies to $q^2 = 2k^2$. This means $q^2$ is also even, and so is $q$. But this contradicts our initial assumption that the fraction $p/q$ was in its simplest form, since both $p$ and $q$ are even and can be divided by 2. This contradiction proves our initial assumption was false, and thus, $\sqrt{2}$ is irrational.
