To prove that there exists a real number $x$ such that $x^2 = 2$, we can use a combination of the Intermediate Value Theorem (IVT) and properties of continuous functions. Here's a detailed proof:

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Step 1: Define the function

Let $f(x) = x^2 - 2$. We are looking for a real number $x$ such that $f(x) = 0$. In other words, we are solving the equation $x^2 = 2$.

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Step 2: Check continuity

The function $f(x) = x^2 - 2$ is a polynomial and hence continuous everywhere on $\mathbb{R}$.

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Step 3: Apply the Intermediate Value Theorem

The IVT states that if $f$ is continuous on a closed interval $[a, b]$, and if $f(a)$ and $f(b)$ have opposite signs, then there exists at least one $c \in (a, b)$ such that $f(c) = 0$.

1. Evaluate $f(x)$ at two points:

   • $f(1) = 1^2 - 2 = -1$
   • $f(2) = 2^2 - 2 = 2$

2. Observe that $f(1) < 0$ and $f(2) > 0$.

3. Since $f$ is continuous on $[1, 2]$ and changes sign between $f(1)$ and $f(2)$, the IVT guarantees that there exists some $c \in (1, 2)$ such that $f(c) = 0$.

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Step 4: Conclude the existence of $x$

The real number $c$ obtained from the IVT satisfies $f(c) = 0$, meaning $c^2 = 2$. Thus, $x = c$ is the real number we were looking for.

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Further Notes

The number $x$ is precisely $\sqrt{2}$, the positive square root of 2, which exists as a real number and is irrational.

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Would you like to explore more details or ask any questions? 😊

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Relative Questions

1. What is the Intermediate Value Theorem, and how does it work?
2. Why is $\sqrt{2}$ considered an irrational number?
3. Can we construct $\sqrt{2}$ geometrically, and how?
4. How can we approximate $\sqrt{2}$ numerically using iterative methods?
5. How does the proof differ for non-real solutions or complex numbers?

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Tip

For any continuous function $f(x)$, the Intermediate Value Theorem is a powerful tool to prove the existence of roots without directly solving the equation!