Prove irrational number

Prompt

How can it be proven that the square root of 2 is an irrational number?

Here is the standard, classical proof by contradiction (sometimes called Euclid’s proof) that √2 cannot be expressed as a ratio of two integers, i.e. that √2 is irrational.

1. Assume the opposite of what we want to prove.  
   Suppose √2 were rational.  Then we could write it as a fraction  p / q, where
   • p and q are integers,  
   • the fraction is in lowest terms (p and q have no common factor other than 1), and  
   • q ≠ 0.

   So we assume
        √2 = p / q.

2. Square both sides to eliminate the square root:
        2 = p² / q².

   Multiply both sides by q²:
        2q² = p².    (★)

3. Conclude that p² is even.  
   From (★), p² equals 2 times q², so p² is divisible by 2, hence p² is even.  
   If p² is even, then p itself must also be even (because the square of an odd number is odd).

   Therefore, we can write p = 2k for some integer k.

4. Substitute p = 2k back into equation (★):
        2q² = (2k)²
        2q² = 4k²
   Divide both sides by 2:
        q² = 2k².

5. Conclude that q² is even, and thus q is even.  
   The equation q² = 2k² shows q² is divisible by 2, so q² is even, which again implies q is even.

6. But p and q cannot both be even.  
   We deduced that both p and q are divisible by 2. That means they share a common factor 2, contradicting our starting assumption that p / q was in lowest terms.

7. Contradiction achieved.  
   Our initial assumption—that √2 can be written as a fraction of two integers in lowest terms—leads to a contradiction. Therefore the assumption must be false.

8. Conclusion.  
   Hence √2 is not rational; it is irrational.

This completes the proof.

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Google

Gemini 2.5 Pro

Of course. The proof that the square root of 2 is an irrational number is one of the most classic and elegant proofs in mathematics. It uses a method called **reductio ad absurdum**, or **proof by contradiction**.

Here is a breakdown of the concept and the step-by-step proof.

### The Core Idea: Proof by Contradiction

The strategy is simple:
1.  We'll start by **assuming the opposite** of what we want to prove. In this case, we'll assume that √2 *is* a rational number.
2.  We will follow the logical steps that stem from this assumption.
3.  We will show that these steps lead to a **logical impossibility**—a contradiction.
4.  Since our initial assumption led to a nonsensical result, the assumption itself must be false. Therefore, the original statement (that √2 is irrational) must be true.

---

### Preliminary Definitions

Before the proof, let's be clear on the terms:

*   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. For example, 0.5 is rational because it's $1/2$. 7 is rational because it's $7/1$.
*   An **irrational number** cannot be expressed as a simple fraction of two integers.
*   A fraction is in its **simplest form** if its numerator and denominator have no common factors other than 1. For example, $2/4$ is not in simplest form, but $1/2$ is. This is the crucial point of the proof.

---

### The Step-by-Step Proof

Here is the formal proof that $\sqrt{2}$ is irrational.

**Step 1: Assume that $\sqrt{2}$ is rational.**
If $\sqrt{2}$ is rational, then by definition, it can be written as a fraction $p/q$ in its simplest form.

$$ \sqrt{2} = \frac{p}{q} $$

Where:
*   *p* and *q* are integers.
*   *q* is not zero ($q \neq 0$).
*   The fraction $p/q$ is in its **simplest form** (p and q have no common factors).

**Step 2: Square both sides of the equation.**
To get rid of the square root, we square both sides:

$$ (\sqrt{2})^2 = \left(\frac{p}{q}\right)^2 $$

$$ 2 = \frac{p^2}{q^2} $$

**Step 3: Rearrange the equation.**
Multiply both sides by $q^2$:

$$ 2q^2 = p^2 $$

**Step 4: Analyze the result.**
This equation, $p^2 = 2q^2$, tells us something important about $p^2$. It says that $p^2$ is equal to 2 multiplied by another integer ($q^2$). This means $p^2$ **must be an even number**.

**Step 5: Deduce a property of *p*.**
If the square of a number ($p^2$) is even, then the number itself (*p*) must also be even.
*(Why? An odd number squared is always odd. For example, $3^2=9$, $5^2=25$. So, if $p^2$ is even, p cannot be odd, which means it must be even).*

**Step 6: Express *p* as an even number.**
Since we've established that *p* is even, we can write it as $2k$, where *k* is some other integer.

$$ p = 2k $$

**Step 7: Substitute this new expression for *p* back into our equation from Step 3.**
The equation was $2q^2 = p^2$. Now, we replace *p* with *2k*:

$$ 2q^2 = (2k)^2 $$

$$ 2q^2 = 4k^2 $$

**Step 8: Simplify the new equation.**
Divide both sides by 2:

$$ q^2 = 2k^2 $$

**Step 9: Analyze this final result.**
Just like in Step 4, this equation tells us that $q^2$ is equal to 2 multiplied by an integer ($k^2$). This means $q^2$ **must also be an even number**.

And, just like in Step 5, if $q^2$ is even, then *q* itself **must also be even**.

### The Contradiction

Let's review what we've just found:
1.  From Step 5, we concluded that **p is even**.
2.  From Step 9, we concluded that **q is even**.

If both *p* and *q* are even, they both have a common factor of 2.

**This is the contradiction!**

Our initial assumption in **Step 1** was that the fraction $p/q$ was in its **simplest form**, meaning *p* and *q* had no common factors. But our logical deduction has shown that they *must* have a common factor of 2.

Both statements cannot be true at the same time. The assumption has led to a logical absurdity.

### Conclusion

Since our initial assumption—that $\sqrt{2}$ is a rational number—leads to a contradiction, the assumption must be false.

Therefore, **the square root of 2 must be an irrational number.**

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OpenAI

gpt-oss-20B (high)

**Proof (by contradiction, “infinite descent”) that \(\sqrt{2}\) is irrational**

1. **Assume the contrary**  
   Suppose \(\sqrt{2}\) *were* rational.  Then it could be written as a fraction of two positive integers in lowest terms:
   \[
   \sqrt{2}=\frac{p}{q}, \qquad \gcd(p,q)=1,\; q>0 .
   \]

2. **Square both sides**  
   \[
   2=\frac{p^{2}}{q^{2}}
   \quad\Longrightarrow\quad
   p^{2}=2q^{2}.   \tag{1}
   \]

3. **Deduce that \(p\) is even**  
   Equation (1) tells us that \(p^{2}\) is divisible by \(2\).  
   The square of an odd integer is odd, so the only way \(p^{2}\) is even is if \(p\) itself is even.  
   Hence we can write
   \[
   p = 2k
   \]
   for some integer \(k\).

4. **Substitute back into (1)**  
   \[
   (2k)^{2} = 2q^{2}
   \quad\Longrightarrow\quad
   4k^{2} = 2q^{2}
   \;\Longrightarrow\;
   q^{2} = 2k^{2}.   \tag{2}
   \]

5. **Deduce that \(q\) is even**  
   Equation (2) shows that \(q^{2}\) is also divisible by \(2\). By the same logic used in step 3, this means \(q\) must be even.

6. **Contradiction**  
   Both \(p\) and \(q\) are even, so they share the common factor \(2\).  
   But we began with the assumption that \(p/q\) was in lowest terms (\(\gcd(p,q)=1\)). Thus our assumption leads to a contradiction.

7. **Conclusion**  
   The contradiction shows that our initial assumption was false. Therefore

\[
\boxed{\;\sqrt{2}\ \text{is not a rational number}\;}
\]

---

### Why the argument works

- **Parity of squares:** An odd integer squared remains odd; an even integer squared is divisible by \(4\). Hence a square being even guarantees that its root is even.
- **Lowest‑term assumption:** By choosing \(p/q\) in lowest terms, we used the fact that two numbers cannot be simultaneously divisible by the same prime (here, \(2\)) unless that prime actually divides the greatest common divisor.
- **Infinite descent idea:** If \(\sqrt{2}\) were rational, we would end up with a smaller pair \((k, q/2)\) that still satisfies the same relation \(p^{2}=2q^{2}\), repeating the reasoning indefinitely. In finite arithmetic this is impossible.

This method generalises: if a positive integer \(n\) is *not* a perfect square, then \(\sqrt{n}\) must be irrational. The proof for \(\sqrt{2}\) is the simplest illustration.

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OpenAI

GPT-4.1

Certainly! To **prove that the square root of 2 (\(\sqrt{2}\)) is an irrational number**, we'll use a classic *proof by contradiction*. Here's the step-by-step reasoning:

---

### 1. **Assume the Opposite**

Suppose \(\sqrt{2}\) *is* rational.  
That means it can be written as a fraction of two integers, \(\frac{a}{b}\), with \(a, b\) having **no common factors** (other than 1), and \(b \neq 0\).

So:  
\[
\sqrt{2} = \frac{a}{b}
\]

### 2. **Square Both Sides**

\[
(\sqrt{2})^2 = \left(\frac{a}{b}\right)^2
\]
\[
2 = \frac{a^2}{b^2}
\]

### 3. **Multiply Both Sides by \(b^2\)**

\[
2b^2 = a^2
\]

So, \(a^2\) is **even** (because it equals \(2\) times some integer).

---

### 4. **Show \(a\) Must Be Even**

If \(a^2\) is even, then \(a\) itself must be even (since the square of an odd number is odd).

Let’s write \(a = 2k\), where \(k\) is an integer.

---

### 5. **Substitute Back**

Substitute \(a = 2k\) into the equation \(2b^2 = a^2\):

\[
2b^2 = (2k)^2
\]
\[
2b^2 = 4k^2
\]
\[
b^2 = 2k^2
\]

So, **\(b^2\) is also even,** and so \(b\) is even as well.

---

### 6. **Contradiction**

If **both** \(a\) and \(b\) are even, they both have 2 as a factor.  
But we **assumed** \(\frac{a}{b}\) is in lowest terms (no common factors).

**This is a contradiction.**

---

### 7. **Conclusion**

Our assumption that \(\sqrt{2}\) could be written as a rational fraction leads to a contradiction.

**Therefore:**

\[
\boxed{\sqrt{2} \text{ is irrational.}}
\]

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Alibaba

Qwen3 235B A22B (Non-reasoning)

To prove that the square root of 2 is an irrational number, we use a classic proof by contradiction, which is one of the most elegant and historically significant arguments in mathematics.

---

### **Understanding the Goal**

We want to show that the number $ \sqrt{2} $ is **irrational**, meaning it cannot be expressed as a ratio of two integers $ \frac{a}{b} $, where $ a $ and $ b $ are integers with no common factors (i.e., the fraction is in lowest terms).

---

### **Step-by-Step Proof by Contradiction**

1. **Assumption (for contradiction):**  
   Suppose $ \sqrt{2} $ is rational. Then, there exist integers $ a $ and $ b $ (with $ b \neq 0 $) such that  
   $$
   \sqrt{2} = \frac{a}{b}
   $$  
   where $ a $ and $ b $ have no common factors (i.e., the fraction is in lowest terms).

2. **Square both sides:**  
   $$
   (\sqrt{2})^2 = \frac{a^2}{b^2} \Rightarrow 2 = \frac{a^2}{b^2}
   $$

3. **Multiply both sides by $ b^2 $:**  
   $$
   2b^2 = a^2
   $$

4. **Analyze the resulting equation $ 2b^2 = a^2 $:**  
   This shows that $ a^2 $ is an even number (since it's equal to twice something), so $ a $ must also be even.  
   (Note: The square of an odd number is odd, while the square of an even number is even.)

5. **Express $ a $ as $ 2k $ for some integer $ k $:**  
   Let $ a = 2k $. Then:  
   $$
   a^2 = (2k)^2 = 4k^2
   $$

6. **Substitute into the equation:**  
   $$
   2b^2 = 4k^2 \Rightarrow b^2 = 2k^2
   $$

7. **Analyze this new equation $ b^2 = 2k^2 $:**  
   This shows that $ b^2 $ is also even, so $ b $ must be even.

8. **Contradiction arises:**  
   If both $ a $ and $ b $ are even, then they share a common factor of 2. This contradicts our original assumption that $ a $ and $ b $ have no common factors.

---

### **Conclusion**

Since our assumption that $ \sqrt{2} $ is rational leads to a contradiction, we conclude that $ \sqrt{2} $ **cannot** be expressed as a ratio of two integers. Therefore, $ \sqrt{2} $ is **irrational**.

This proof is a cornerstone of number theory and has been known since antiquity, attributed to the Pythagoreans. It elegantly illustrates why a simple geometric object — the diagonal of a unit square — leads to an irrational number, fundamentally altering our understanding of numbers.

---

### **Final Answer**

$$
\boxed{\sqrt{2}}
$$

is irrational.

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DeepSeek

DeepSeek V3 0324

To prove that the square root of 2 (\(\sqrt{2}\)) is an **irrational number**, we can use a classic method known as **proof by contradiction**. 

### Proof:
1. **Assumption for Contradiction:**
   - Assume that \(\sqrt{2}\) is a **rational number**. 
   - By definition, a rational number can be expressed as a fraction \(\frac{a}{b}\), where \(a\) and \(b\) are integers (with \(b \neq 0\)) and the fraction is in its **simplest form** (i.e., \(a\) and \(b\) have no common divisors other than 1).

2. **Express \(\sqrt{2}\) as a Fraction:**
   \[
   \sqrt{2} = \frac{a}{b}
   \]
   - Square both sides:
   \[
   2 = \frac{a^2}{b^2}
   \]
   - Multiply both sides by \(b^2\):
   \[
   2b^2 = a^2
   \]
   - This implies that \(a^2\) is **even** (since it is equal to \(2\) times an integer).

3. **Consequence for \(a\):**
   - If \(a^2\) is even, then \(a\) itself must be even (since the square of an odd number is odd).
   - Let \(a = 2k\), where \(k\) is an integer.
   - Substitute back into the equation:
   \[
   2b^2 = (2k)^2 \implies 2b^2 = 4k^2
   \]
   - Divide both sides by 2:
   \[
   b^2 = 2k^2
   \]
   - This implies that \(b^2\) is also even, and hence \(b\) is even.

4. **Contradiction:**
   - If both \(a\) and \(b\) are even, they have a common divisor of 2.
   - This contradicts our initial assumption that \(\frac{a}{b}\) is in its **simplest form** (with no common divisors other than 1).

5. **Conclusion:**
   - Our original assumption that \(\sqrt{2}\) is rational must be **false**.
   - Therefore, \(\sqrt{2}\) is **irrational**.

\[
\boxed{\sqrt{2} \text{ is irrational}}
\]

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Meta

Llama 4 Maverick

## Step 1: Understand the problem and the definition of an irrational number
To prove that the square root of 2 is an irrational number, we first need to understand what an irrational number is. An irrational number is a number that cannot be expressed as a finite decimal or fraction. It has an infinite number of digits, and these digits never repeat in a predictable pattern.

## Step 2: Assume the square root of 2 is rational
Let's assume, for the sake of contradiction, that the square root of 2 is a rational number. This means it can be expressed as a fraction, where the numerator and denominator are integers and the denominator is non-zero. We can further assume that this fraction is in its simplest form, meaning the numerator and denominator have no common factors other than 1.

## Step 3: Express the square root of 2 as a fraction
Given our assumption, we can write $\sqrt{2} = \frac{a}{b}$, where $a$ and $b$ are integers and $\frac{a}{b}$ is in its simplest form. This implies that $a$ and $b$ have no common factors other than 1.

## Step 4: Square both sides of the equation
Squaring both sides gives $2 = \frac{a^2}{b^2}$. This equation can be rearranged to $a^2 = 2b^2$.

## Step 5: Analyze the equation for $a^2 = 2b^2$
From $a^2 = 2b^2$, we can deduce that $a^2$ is an even number because it is equal to 2 times $b^2$, and $b^2$ is an integer. If $a^2$ is even, then $a$ must also be even because the square of an odd number is always odd.

## Step 6: Express $a$ as $2k$
Since $a$ is even, we can express $a$ as $2k$, where $k$ is an integer. Substituting $a = 2k$ into $a^2 = 2b^2$ gives $(2k)^2 = 2b^2$, which simplifies to $4k^2 = 2b^2$.

## Step 7: Simplify the equation to show $b$ is even
Dividing both sides of $4k^2 = 2b^2$ by 2 gives $2k^2 = b^2$. This implies that $b^2$ is even, and therefore $b$ must also be even.

## Step 8: Recognize the contradiction
We have shown that both $a$ and $b$ are even. This means they have a common factor of 2, which contradicts our initial assumption that $\frac{a}{b}$ is in its simplest form, with $a$ and $b$ having no common factors other than 1.

## Step 9: Conclude that the square root of 2 is irrational
Since our assumption that $\sqrt{2}$ is rational leads to a contradiction, it must be false. Therefore, $\sqrt{2}$ is not a rational number; it is an irrational number.

The final answer is: $\boxed{There is no number to solve this problem. The task was to prove that $\sqrt{2}$ is irrational, which has been demonstrated through the steps above.}$

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DeepSeek

DeepSeek R1 0528 (May '25)

To prove that \(\sqrt{2}\) is irrational, we use a proof by contradiction. Here is the step-by-step reasoning:

1. **Assumption**: Assume \(\sqrt{2}\) is rational. This means it can be expressed as a fraction \(\frac{a}{b}\) in its lowest terms, where \(a\) and \(b\) are integers with no common divisors other than 1 (i.e., \(\gcd(a, b) = 1\)) and \(b \neq 0\).

2. **Equation from the assumption**: Since \(\sqrt{2} = \frac{a}{b}\), squaring both sides gives:  
   \[
   2 = \frac{a^2}{b^2}
   \]  
   Rearranging this, we get:  
   \[
   a^2 = 2b^2
   \]

3. **Deduction about \(a\)**: The equation \(a^2 = 2b^2\) implies that \(a^2\) is even (since it is twice \(b^2\)). If \(a^2\) is even, then \(a\) must also be even. This is because the square of an odd number is always odd, so an even square must come from an even number. Therefore, we can write \(a = 2k\) for some integer \(k\).

4. **Substitution**: Substitute \(a = 2k\) into the equation \(a^2 = 2b^2\):  
   \[
   (2k)^2 = 2b^2 \implies 4k^2 = 2b^2 \implies 2k^2 = b^2
   \]

5. **Deduction about \(b\)**: The equation \(b^2 = 2k^2\) implies that \(b^2\) is even. Similarly, if \(b^2\) is even, then \(b\) must be even. Therefore, \(b\) is even.

6. **Contradiction**: From steps 3 and 5, both \(a\) and \(b\) are even. This means both are divisible by 2, contradicting the initial assumption that \(\frac{a}{b}\) is in its lowest terms (i.e., that \(\gcd(a, b) = 1\)).

Since the assumption that \(\sqrt{2}\) is rational leads to a contradiction, it must be false. Therefore, \(\sqrt{2}\) is irrational.

\boxed{\text{By assuming } \sqrt{2} \text{ is rational and expressing it as } \frac{a}{b} \text{ in lowest terms, we find that both } a \text{ and } b \text{ must be even, contradicting coprimality. Thus, } \sqrt{2} \text{ is irrational.}}

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Anthropic

Claude 4 Sonnet (Non-reasoning)

The most elegant proof that √2 is irrational uses **proof by contradiction**. Here's how it works:

## The Proof

**Assumption**: Suppose √2 is rational (we'll show this leads to a contradiction)

**Step 1**: If √2 is rational, then it can be written as:
√2 = a/b

where a and b are integers with no common factors (the fraction is in lowest terms), and b ≠ 0.

**Step 2**: Square both sides:
2 = a²/b²

**Step 3**: Multiply both sides by b²:
2b² = a²

**Step 4**: Since a² = 2b², we know that a² is even (it's 2 times something). 
If a² is even, then a must be even. (If a were odd, then a² would be odd.)

**Step 5**: Since a is even, we can write a = 2k for some integer k.
Substituting back: 2b² = (2k)² = 4k²

**Step 6**: Dividing by 2: b² = 2k²

**Step 7**: Now b² is even (it equals 2k²), which means b is also even.

**The Contradiction**: We've shown that both a and b must be even, meaning they share the common factor 2. But this contradicts our assumption in Step 1 that a/b was in lowest terms (no common factors).

**Conclusion**: Since our assumption led to a contradiction, √2 cannot be rational. Therefore, √2 is irrational.

This proof is attributed to the ancient Greeks and demonstrates the power of logical reasoning in mathematics!

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Here is the standard, classical proof by contradiction (sometimes called Euclid’s proof) that √2 cannot be expressed as a ratio of two integers, i.e. that √2 is irrational. 1. Assume the opposite of what we want to prove. Suppose √2 were rational. Then we could write it as a fraction p / q, where • p and q are integers, • the fraction is in lowest terms (p and q have no common factor other than 1), and • q ≠ 0. So we assume √2 = p / q. 2. Square both sides to eliminate the square root: 2 = p² / q². Multiply both sides by q²: 2q² = p². (★) 3. Conclude that p² is even. From (★), p² equals 2 times q², so p² is divisible by 2, hence p² is even. If p² is even, then p itself must also be even (because the square of an odd number is odd). Therefore, we can write p = 2k for some integer k. 4. Substitute p = 2k back into equation (★): 2q² = (2k)² 2q² = 4k² Divide both sides by 2: q² = 2k². 5. Conclude that q² is even, and thus q is even. The equation q² = 2k² shows q² is divisible by 2, so q² is even, which again implies q is even. 6. But p and q cannot both be even. We deduced that both p and q are divisible by 2. That means they share a common factor 2, contradicting our starting assumption that p / q was in lowest terms. 7. Contradiction achieved. Our initial assumption—that √2 can be written as a fraction of two integers in lowest terms—leads to a contradiction. Therefore the assumption must be false. 8. Conclusion. Hence √2 is not rational; it is irrational. This completes the proof.

Of course. The proof that the square root of 2 is an irrational number is one of the most classic and elegant proofs in mathematics. It uses a method called **reductio ad absurdum**, or **proof by contradiction**. Here is a breakdown of the concept and the step-by-step proof. ### The Core Idea: Proof by Contradiction The strategy is simple: 1. We'll start by **assuming the opposite** of what we want to prove. In this case, we'll assume that √2 *is* a rational number. 2. We will follow the logical steps that stem from this assumption. 3. We will show that these steps lead to a **logical impossibility**—a contradiction. 4. Since our initial assumption led to a nonsensical result, the assumption itself must be false. Therefore, the original statement (that √2 is irrational) must be true. --- ### Preliminary Definitions Before the proof, let's be clear on the terms: * 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. For example, 0.5 is rational because it's $1/2$. 7 is rational because it's $7/1$. * An **irrational number** cannot be expressed as a simple fraction of two integers. * A fraction is in its **simplest form** if its numerator and denominator have no common factors other than 1. For example, $2/4$ is not in simplest form, but $1/2$ is. This is the crucial point of the proof. --- ### The Step-by-Step Proof Here is the formal proof that $\sqrt{2}$ is irrational. **Step 1: Assume that $\sqrt{2}$ is rational.** If $\sqrt{2}$ is rational, then by definition, it can be written as a fraction $p/q$ in its simplest form. $$ \sqrt{2} = \frac{p}{q} $$ Where: * *p* and *q* are integers. * *q* is not zero ($q \neq 0$). * The fraction $p/q$ is in its **simplest form** (p and q have no common factors). **Step 2: Square both sides of the equation.** To get rid of the square root, we square both sides: $$ (\sqrt{2})^2 = \left(\frac{p}{q}\right)^2 $$ $$ 2 = \frac{p^2}{q^2} $$ **Step 3: Rearrange the equation.** Multiply both sides by $q^2$: $$ 2q^2 = p^2 $$ **Step 4: Analyze the result.** This equation, $p^2 = 2q^2$, tells us something important about $p^2$. It says that $p^2$ is equal to 2 multiplied by another integer ($q^2$). This means $p^2$ **must be an even number**. **Step 5: Deduce a property of *p*.** If the square of a number ($p^2$) is even, then the number itself (*p*) must also be even. *(Why? An odd number squared is always odd. For example, $3^2=9$, $5^2=25$. So, if $p^2$ is even, p cannot be odd, which means it must be even).* **Step 6: Express *p* as an even number.** Since we've established that *p* is even, we can write it as $2k$, where *k* is some other integer. $$ p = 2k $$ **Step 7: Substitute this new expression for *p* back into our equation from Step 3.** The equation was $2q^2 = p^2$. Now, we replace *p* with *2k*: $$ 2q^2 = (2k)^2 $$ $$ 2q^2 = 4k^2 $$ **Step 8: Simplify the new equation.** Divide both sides by 2: $$ q^2 = 2k^2 $$ **Step 9: Analyze this final result.** Just like in Step 4, this equation tells us that $q^2$ is equal to 2 multiplied by an integer ($k^2$). This means $q^2$ **must also be an even number**. And, just like in Step 5, if $q^2$ is even, then *q* itself **must also be even**. ### The Contradiction Let's review what we've just found: 1. From Step 5, we concluded that **p is even**. 2. From Step 9, we concluded that **q is even**. If both *p* and *q* are even, they both have a common factor of 2. **This is the contradiction!** Our initial assumption in **Step 1** was that the fraction $p/q$ was in its **simplest form**, meaning *p* and *q* had no common factors. But our logical deduction has shown that they *must* have a common factor of 2. Both statements cannot be true at the same time. The assumption has led to a logical absurdity. ### Conclusion Since our initial assumption—that $\sqrt{2}$ is a rational number—leads to a contradiction, the assumption must be false. Therefore, **the square root of 2 must be an irrational number.**

**Proof (by contradiction, “infinite descent”) that $\sqrt{2}$ is irrational** 1. **Assume the contrary** Suppose $\sqrt{2}$ *were* rational. Then it could be written as a fraction of two positive integers in lowest terms: \[ \sqrt{2}=\frac{p}{q}, \qquad \gcd(p,q)=1,\; q>0 . \] 2. **Square both sides** \[ 2=\frac{p^{2}}{q^{2}} \quad\Longrightarrow\quad p^{2}=2q^{2}. \tag{1} \] 3. **Deduce that $p$ is even** Equation (1) tells us that $p^{2}$ is divisible by $2$. The square of an odd integer is odd, so the only way $p^{2}$ is even is if $p$ itself is even. Hence we can write \[ p = 2k \] for some integer $k$. 4. **Substitute back into (1)** \[ (2k)^{2} = 2q^{2} \quad\Longrightarrow\quad 4k^{2} = 2q^{2} \;\Longrightarrow\; q^{2} = 2k^{2}. \tag{2} \] 5. **Deduce that $q$ is even** Equation (2) shows that $q^{2}$ is also divisible by $2$. By the same logic used in step 3, this means $q$ must be even. 6. **Contradiction** Both $p$ and $q$ are even, so they share the common factor $2$. But we began with the assumption that $p/q$ was in lowest terms ($\gcd(p,q)=1$). Thus our assumption leads to a contradiction. 7. **Conclusion** The contradiction shows that our initial assumption was false. Therefore \[ \boxed{\;\sqrt{2}\ \text{is not a rational number}\;} \] --- ### Why the argument works - **Parity of squares:** An odd integer squared remains odd; an even integer squared is divisible by $4$. Hence a square being even guarantees that its root is even. - **Lowest‑term assumption:** By choosing $p/q$ in lowest terms, we used the fact that two numbers cannot be simultaneously divisible by the same prime (here, $2$) unless that prime actually divides the greatest common divisor. - **Infinite descent idea:** If $\sqrt{2}$ were rational, we would end up with a smaller pair $(k, q/2)$ that still satisfies the same relation $p^{2}=2q^{2}$, repeating the reasoning indefinitely. In finite arithmetic this is impossible. This method generalises: if a positive integer $n$ is *not* a perfect square, then $\sqrt{n}$ must be irrational. The proof for $\sqrt{2}$ is the simplest illustration.

Certainly! To **prove that the square root of 2 ($\sqrt{2}$) is an irrational number**, we'll use a classic *proof by contradiction*. Here's the step-by-step reasoning: --- ### 1. **Assume the Opposite** Suppose $\sqrt{2}$ *is* rational. That means it can be written as a fraction of two integers, $\frac{a}{b}$, with $a, b$ having **no common factors** (other than 1), and $b \neq 0$. So: \[ \sqrt{2} = \frac{a}{b} \] ### 2. **Square Both Sides** \[ (\sqrt{2})^2 = \left(\frac{a}{b}\right)^2 \] \[ 2 = \frac{a^2}{b^2} \] ### 3. **Multiply Both Sides by $b^2$** \[ 2b^2 = a^2 \] So, $a^2$ is **even** (because it equals $2$ times some integer). --- ### 4. **Show $a$ Must Be Even** If $a^2$ is even, then $a$ itself must be even (since the square of an odd number is odd). Let’s write $a = 2k$, where $k$ is an integer. --- ### 5. **Substitute Back** Substitute $a = 2k$ into the equation $2b^2 = a^2$: \[ 2b^2 = (2k)^2 \] \[ 2b^2 = 4k^2 \] \[ b^2 = 2k^2 \] So, **$b^2$ is also even,** and so $b$ is even as well. --- ### 6. **Contradiction** If **both** $a$ and $b$ are even, they both have 2 as a factor. But we **assumed** $\frac{a}{b}$ is in lowest terms (no common factors). **This is a contradiction.** --- ### 7. **Conclusion** Our assumption that $\sqrt{2}$ could be written as a rational fraction leads to a contradiction. **Therefore:** \[ \boxed{\sqrt{2} \text{ is irrational.}} \]

To prove that the square root of 2 is an irrational number, we use a classic proof by contradiction, which is one of the most elegant and historically significant arguments in mathematics. --- ### **Understanding the Goal** We want to show that the number $ \sqrt{2} $ is **irrational**, meaning it cannot be expressed as a ratio of two integers $ \frac{a}{b} $, where $ a $ and $ b $ are integers with no common factors (i.e., the fraction is in lowest terms). --- ### **Step-by-Step Proof by Contradiction** 1. **Assumption (for contradiction):** Suppose $ \sqrt{2} $ is rational. Then, there exist integers $ a $ and $ b $ (with $ b \neq 0 $) such that $$ \sqrt{2} = \frac{a}{b} $$ where $ a $ and $ b $ have no common factors (i.e., the fraction is in lowest terms). 2. **Square both sides:** $$ (\sqrt{2})^2 = \frac{a^2}{b^2} \Rightarrow 2 = \frac{a^2}{b^2} $$ 3. **Multiply both sides by $ b^2 $:** $$ 2b^2 = a^2 $$ 4. **Analyze the resulting equation $ 2b^2 = a^2 $:** This shows that $ a^2 $ is an even number (since it's equal to twice something), so $ a $ must also be even. (Note: The square of an odd number is odd, while the square of an even number is even.) 5. **Express $ a $ as $ 2k $ for some integer $ k $:** Let $ a = 2k $. Then: $$ a^2 = (2k)^2 = 4k^2 $$ 6. **Substitute into the equation:** $$ 2b^2 = 4k^2 \Rightarrow b^2 = 2k^2 $$ 7. **Analyze this new equation $ b^2 = 2k^2 $:** This shows that $ b^2 $ is also even, so $ b $ must be even. 8. **Contradiction arises:** If both $ a $ and $ b $ are even, then they share a common factor of 2. This contradicts our original assumption that $ a $ and $ b $ have no common factors. --- ### **Conclusion** Since our assumption that $ \sqrt{2} $ is rational leads to a contradiction, we conclude that $ \sqrt{2} $ **cannot** be expressed as a ratio of two integers. Therefore, $ \sqrt{2} $ is **irrational**. This proof is a cornerstone of number theory and has been known since antiquity, attributed to the Pythagoreans. It elegantly illustrates why a simple geometric object — the diagonal of a unit square — leads to an irrational number, fundamentally altering our understanding of numbers. --- ### **Final Answer** $$ \boxed{\sqrt{2}} $$ is irrational.

To prove that the square root of 2 ($\sqrt{2}$) is an **irrational number**, we can use a classic method known as **proof by contradiction**. ### Proof: 1. **Assumption for Contradiction:** - Assume that $\sqrt{2}$ is a **rational number**. - By definition, a rational number can be expressed as a fraction $\frac{a}{b}$, where $a$ and $b$ are integers (with $b \neq 0$) and the fraction is in its **simplest form** (i.e., $a$ and $b$ have no common divisors other than 1). 2. **Express $\sqrt{2}$ as a Fraction:** \[ \sqrt{2} = \frac{a}{b} \] - Square both sides: \[ 2 = \frac{a^2}{b^2} \] - Multiply both sides by $b^2$: \[ 2b^2 = a^2 \] - This implies that $a^2$ is **even** (since it is equal to $2$ times an integer). 3. **Consequence for $a$:** - If $a^2$ is even, then $a$ itself must be even (since the square of an odd number is odd). - Let $a = 2k$, where $k$ is an integer. - Substitute back into the equation: \[ 2b^2 = (2k)^2 \implies 2b^2 = 4k^2 \] - Divide both sides by 2: \[ b^2 = 2k^2 \] - This implies that $b^2$ is also even, and hence $b$ is even. 4. **Contradiction:** - If both $a$ and $b$ are even, they have a common divisor of 2. - This contradicts our initial assumption that $\frac{a}{b}$ is in its **simplest form** (with no common divisors other than 1). 5. **Conclusion:** - Our original assumption that $\sqrt{2}$ is rational must be **false**. - Therefore, $\sqrt{2}$ is **irrational**. \[ \boxed{\sqrt{2} \text{ is irrational}} \]

## Step 1: Understand the problem and the definition of an irrational number To prove that the square root of 2 is an irrational number, we first need to understand what an irrational number is. An irrational number is a number that cannot be expressed as a finite decimal or fraction. It has an infinite number of digits, and these digits never repeat in a predictable pattern. ## Step 2: Assume the square root of 2 is rational Let's assume, for the sake of contradiction, that the square root of 2 is a rational number. This means it can be expressed as a fraction, where the numerator and denominator are integers and the denominator is non-zero. We can further assume that this fraction is in its simplest form, meaning the numerator and denominator have no common factors other than 1. ## Step 3: Express the square root of 2 as a fraction Given our assumption, we can write $\sqrt{2} = \frac{a}{b}$, where $a$ and $b$ are integers and $\frac{a}{b}$ is in its simplest form. This implies that $a$ and $b$ have no common factors other than 1. ## Step 4: Square both sides of the equation Squaring both sides gives $2 = \frac{a^2}{b^2}$. This equation can be rearranged to $a^2 = 2b^2$. ## Step 5: Analyze the equation for $a^2 = 2b^2$ From $a^2 = 2b^2$, we can deduce that $a^2$ is an even number because it is equal to 2 times $b^2$, and $b^2$ is an integer. If $a^2$ is even, then $a$ must also be even because the square of an odd number is always odd. ## Step 6: Express $a$ as $2k$ Since $a$ is even, we can express $a$ as $2k$, where $k$ is an integer. Substituting $a = 2k$ into $a^2 = 2b^2$ gives $(2k)^2 = 2b^2$, which simplifies to $4k^2 = 2b^2$. ## Step 7: Simplify the equation to show $b$ is even Dividing both sides of $4k^2 = 2b^2$ by 2 gives $2k^2 = b^2$. This implies that $b^2$ is even, and therefore $b$ must also be even. ## Step 8: Recognize the contradiction We have shown that both $a$ and $b$ are even. This means they have a common factor of 2, which contradicts our initial assumption that $\frac{a}{b}$ is in its simplest form, with $a$ and $b$ having no common factors other than 1. ## Step 9: Conclude that the square root of 2 is irrational Since our assumption that $\sqrt{2}$ is rational leads to a contradiction, it must be false. Therefore, $\sqrt{2}$ is not a rational number; it is an irrational number. The final answer is: $\boxed{There is no number to solve this problem. The task was to prove that $\sqrt{2}$ is irrational, which has been demonstrated through the steps above.}$

To prove that $\sqrt{2}$ is irrational, we use a proof by contradiction. Here is the step-by-step reasoning: 1. **Assumption**: Assume $\sqrt{2}$ is rational. This means it can be expressed as a fraction $\frac{a}{b}$ in its lowest terms, where $a$ and $b$ are integers with no common divisors other than 1 (i.e., $\gcd(a, b) = 1$) and $b \neq 0$. 2. **Equation from the assumption**: Since $\sqrt{2} = \frac{a}{b}$, squaring both sides gives: \[ 2 = \frac{a^2}{b^2} \] Rearranging this, we get: \[ a^2 = 2b^2 \] 3. **Deduction about $a$**: The equation $a^2 = 2b^2$ implies that $a^2$ is even (since it is twice $b^2$). If $a^2$ is even, then $a$ must also be even. This is because the square of an odd number is always odd, so an even square must come from an even number. Therefore, we can write $a = 2k$ for some integer $k$. 4. **Substitution**: Substitute $a = 2k$ into the equation $a^2 = 2b^2$: \[ (2k)^2 = 2b^2 \implies 4k^2 = 2b^2 \implies 2k^2 = b^2 \] 5. **Deduction about $b$**: The equation $b^2 = 2k^2$ implies that $b^2$ is even. Similarly, if $b^2$ is even, then $b$ must be even. Therefore, $b$ is even. 6. **Contradiction**: From steps 3 and 5, both $a$ and $b$ are even. This means both are divisible by 2, contradicting the initial assumption that $\frac{a}{b}$ is in its lowest terms (i.e., that $\gcd(a, b) = 1$). Since the assumption that $\sqrt{2}$ is rational leads to a contradiction, it must be false. Therefore, $\sqrt{2}$ is irrational. \boxed{\text{By assuming } \sqrt{2} \text{ is rational and expressing it as } \frac{a}{b} \text{ in lowest terms, we find that both } a \text{ and } b \text{ must be even, contradicting coprimality. Thus, } \sqrt{2} \text{ is irrational.}}

The most elegant proof that √2 is irrational uses **proof by contradiction**. Here's how it works: ## The Proof **Assumption**: Suppose √2 is rational (we'll show this leads to a contradiction) **Step 1**: If √2 is rational, then it can be written as: √2 = a/b where a and b are integers with no common factors (the fraction is in lowest terms), and b ≠ 0. **Step 2**: Square both sides: 2 = a²/b² **Step 3**: Multiply both sides by b²: 2b² = a² **Step 4**: Since a² = 2b², we know that a² is even (it's 2 times something). If a² is even, then a must be even. (If a were odd, then a² would be odd.) **Step 5**: Since a is even, we can write a = 2k for some integer k. Substituting back: 2b² = (2k)² = 4k² **Step 6**: Dividing by 2: b² = 2k² **Step 7**: Now b² is even (it equals 2k²), which means b is also even. **The Contradiction**: We've shown that both a and b must be even, meaning they share the common factor 2. But this contradicts our assumption in Step 1 that a/b was in lowest terms (no common factors). **Conclusion**: Since our assumption led to a contradiction, √2 cannot be rational. Therefore, √2 is irrational. This proof is attributed to the ancient Greeks and demonstrates the power of logical reasoning in mathematics!