Next, we square both sides of the equation √2 = (a)/(b) to eliminate the square root. This gives us 2 = (a²)/(b²). Rearranging the terms, we get a² = 2b². This equation tells us that a² is an even number, because it is equal to 2 times another integer (b²).

Next, we square both sides of the equation √2 = (a)/(b) to eliminate the square root. This gives us 2 = (a²)/(b²). Rearranging the terms, we get a² = 2b². This equation tells us that a² is an even number, because it is equal to 2 times another integer (b²).

If a² is an even number, then a itself must also be an even number. This is because the square of an odd number is always odd ((2n+1)² = 4n²+4n+1 = 2(2n²+2n)+1), and the square of an even number is always even ((2n)² = 4n² = 2(2n²)). Therefore, we can write a as 2k for some integer k.

If a² is an even number, then a itself must also be an even number. This is because the square of an odd number is always odd ((2n+1)² = 4n²+4n+1 = 2(2n²+2n)+1), and the square of an even number is always even ((2n)² = 4n² = 2(2n²)). Therefore, we can write a as 2k for some integer k.

Now we substitute a = 2k back into our equation a² = 2b². This gives us (2k)² = 2b², which simplifies to 4k² = 2b². Dividing both sides by 2, we get 2k² = b². This equation shows that b² is an even number, because it is equal to 2 times another integer (k²).

Now we substitute a = 2k back into our equation a² = 2b². This gives us (2k)² = 2b², which simplifies to 4k² = 2b². Dividing both sides by 2, we get 2k² = b². This equation shows that b² is an even number, because it is equal to 2 times another integer (k²).

Since our initial assumption that √2 is rational led to a contradiction, we must conclude that the assumption is false. Therefore, √2 cannot be expressed as a fraction of two integers and is, by definition, an irrational number.

Since our initial assumption that √2 is rational led to a contradiction, we must conclude that the assumption is false. Therefore, √2 cannot be expressed as a fraction of two integers and is, by definition, an irrational number.

Prove that the square root of 2 is irrational.

Answer: $\sqrt{2}$ is irrational.

Step-by-step solution

Step 1: Assume $\sqrt{2}$ is rational

To prove that $\sqrt{2}$ is irrational, we will use the method of proof by contradiction. We start by assuming the opposite: 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, $b \neq 0$ , and $a$ and $b$ have no common factors other than 1 (i.e., the fraction is in its simplest form).

Step 2: Square both sides

Next, we square both sides of the equation $\sqrt{2} = \frac{a}{b}$ to eliminate the square root. This gives us $2 = \frac{a^2}{b^2}$ . Rearranging the terms, we get $a^2 = 2b^2$ . This equation tells us that $a^2$ is an even number, because it is equal to 2 times another integer ( $b^2$ ).

Step 3: Deduce $a$ is even

If $a^2$ is an even number, then $a$ itself must also be an even number. This is because the square of an odd number is always odd ( $(2n+1)^2 = 4n^2+4n+1 = 2(2n^2+2n)+1$ ), and the square of an even number is always even ( $(2n)^2 = 4n^2 = 2(2n^2)$ ). Therefore, we can write $a$ as $2k$ for some integer $k$ .

Step 4: Substitute $a$ into the equation

Now we substitute $a = 2k$ back into our equation $a^2 = 2b^2$ . This gives us $(2k)^2 = 2b^2$ , which simplifies to $4k^2 = 2b^2$ . Dividing both sides by 2, we get $2k^2 = b^2$ . This equation shows that $b^2$ is an even number, because it is equal to 2 times another integer ( $k^2$ ).

Step 5: Deduce $b$ is even and find contradiction

Similar to the previous step, if $b^2$ is an even number, then $b$ itself must also be an even number. So, we can write $b$ as $2m$ for some integer $m$ . This means that both $a$ and $b$ are even numbers, which implies they both have a common factor of 2. However, this contradicts our initial assumption that $a$ and $b$ have no common factors other than 1. Since our initial assumption leads to a contradiction, it must be false.

Step 6: Conclusion

Since our initial assumption that $\sqrt{2}$ is rational led to a contradiction, we must conclude that the assumption is false. Therefore, $\sqrt{2}$ cannot be expressed as a fraction of two integers and is, by definition, an irrational number.

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Prove that the square root of 2 is irrational.

Step-by-step solution

Step 1: Assume 2\sqrt{2}2​ is rational