In context of discrete math prove that the square root of 2 is irrational number 

To prove that \(\sqrt{2}\) is irrational, we will proceed by *contradiction*. That is, we will assume that \(\sqrt{2}\) is rational and show that this assumption leads to a contradiction.

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### Proof by Contradiction:
1. **Assumption**:  
   Assume \(\sqrt{2}\) is rational. This means that it can be written as a fraction of two integers:
   \[
   \sqrt{2} = \frac{p}{q}
   \]
   where \(p\) and \(q\) are integers with no common factors (i.e., the fraction \(\frac{p}{q}\) is in its simplest form). Also, \(q \neq 0\).

2. **Square both sides**:  
   \[
   2 = \frac{p^2}{q^2}
   \]
   Multiply both sides by \(q^2\) to get:
   \[
   2q^2 = p^2
   \]
   This shows that \(p^2\) is even, since it is equal to \(2q^2\).

3. **Implication that \(p\) is even**:  
   Since \(p^2\) is even, \(p\) must also be even. (If \(p\) were odd, then \(p^2\) would be odd as well, which contradicts our result.)  
   So, let \(p = 2k\) for some integer \(k\).

4. **Substitute \(p = 2k\) into the equation \(2q^2 = p^2\)**:  
   \[
   2q^2 = (2k)^2 = 4k^2
   \]
   Divide both sides by 2:
   \[
   q^2 = 2k^2
   \]
   This shows that \(q^2\) is also even, which means that \(q\) must be even as well.

5. **Contradiction**:  
   Since both \(p\) and \(q\) are even, they have a common factor of 2. This contradicts our assumption that \(\frac{p}{q}\) is in its simplest form (i.e., that \(p\) and \(q\) have no common factors).

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### Conclusion:  
Since our assumption that \(\sqrt{2}\) is rational leads to a contradiction, the original assumption must be false. Therefore, \(\sqrt{2}\) is **irrational**.

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Would you like a deeper breakdown of any part of the proof?  

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Here are 5 related questions for further exploration:
1. How can this proof method be generalized for \(\sqrt{n}\), where \(n\) is not a perfect square?
2. What is the definition of an irrational number, and how does it differ from a rational number?
3. Can similar proofs show that \(\sqrt{3}\) or \(\sqrt{5}\) are irrational?
4. What is the connection between irrational numbers and real numbers?
5. How does this proof relate to the fundamental theorem of arithmetic?

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**Tip:** Proofs by contradiction are powerful tools in mathematics, especially when dealing with irrational numbers and other properties that seem counterintuitive at first.

In context of discrete math prove that the square root of 2 is irrational number

Understand the proof that \(\sqrt{2}\) is irrational with a step-by-step contradiction approach. This problem demonstrates core concepts in number theory and irrational numbers, exploring the fundamentals of proof techniques and properties of rational numbers, ideal for high school students in Grades 10-12.

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