prove that squrt 2 is irrational via contradiction

### Proof by Contradiction: \( \sqrt{2} \) is Irrational

We will prove that \( \sqrt{2} \) is irrational using a **proof by contradiction**. This method assumes the opposite of what we want to prove and then shows that this assumption leads to a contradiction.

#### Step 1: Assume \( \sqrt{2} \) is Rational
Suppose \( \sqrt{2} \) is **rational**. This means we can express \( \sqrt{2} \) as a ratio of two integers in **simplest form**:

\[
\sqrt{2} = \frac{p}{q}
\]

where \( p \) and \( q \) are integers, and the fraction \( \frac{p}{q} \) is in its simplest form (i.e., \( p \) and \( q \) have no common factors other than 1, or they are **coprime**).

#### Step 2: Square Both Sides
Next, we square both sides of the equation to eliminate the square root:

\[
2 = \frac{p^2}{q^2}
\]

Multiplying both sides by \( q^2 \) to clear the denominator:

\[
2q^2 = p^2
\]

#### Step 3: Analyze the Equation \( 2q^2 = p^2 \)
From this equation, we can conclude that \( p^2 \) is **even** (since it is equal to \( 2q^2 \), which is clearly even). Since \( p^2 \) is even, \( p \) itself must be even (because the square of an odd number is odd).

#### Step 4: Express \( p \) as an Even Integer
Since \( p \) is even, we can write \( p = 2k \) for some integer \( k \). Substituting \( p = 2k \) into the equation \( 2q^2 = p^2 \):

\[
2q^2 = (2k)^2
\]

\[
2q^2 = 4k^2
\]

Dividing both sides by 2:

\[
q^2 = 2k^2
\]

#### Step 5: Analyze the Equation \( q^2 = 2k^2 \)
This equation tells us that \( q^2 \) is also even, which implies that \( q \) is even (for the same reason as before: the square of an odd number is odd).

#### Step 6: Contradiction
We have now shown that both \( p \) and \( q \) are even. But this contradicts our assumption that \( \frac{p}{q} \) is in its **simplest form**, because if both \( p \) and \( q \) are even, they share a common factor of 2 and thus the fraction could be simplified further.

#### Step 7: Conclusion
Since our assumption that \( \sqrt{2} \) is rational led to a contradiction, we conclude that \( \sqrt{2} \) cannot be rational. Therefore, \( \sqrt{2} \) is **irrational**.

---

Would you like more details, or do you have any specific questions about this proof?

Here are 5 follow-up questions:
1. Can you generalize this proof to show that \( \sqrt{n} \) is irrational for any non-square \( n \)?
2. How can the concept of irrational numbers be applied to real-life measurements?
3. Why is it important that \( p \) and \( q \) are in simplest form for this proof to work?
4. What other irrational numbers can be proved using a similar method of contradiction?
5. How does this proof relate to the ancient Greeks' discovery of irrational numbers?

**Tip**: When working with proofs, always carefully examine your assumptions—they are often key to identifying contradictions.

prove that sqrt 2 is irrational via contradiction

Explore a detailed proof by contradiction showing that √2 is irrational. This classic problem in number theory involves expressing √2 as a fraction, analyzing even and odd numbers, and concluding with a contradiction. Ideal for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation