Math Problem Statement

prove that root 2 is not rational

Solution

To prove that 2\sqrt{2} is not a rational number, we use a proof by contradiction. Here's the step-by-step process:

Step 1: Assume the Opposite

Assume that 2\sqrt{2} is rational. If 2\sqrt{2} is rational, then it can be expressed as a fraction of two integers aa and bb in simplest form (i.e., the fraction is fully reduced with no common factors other than 1 between aa and bb):

2=ab\sqrt{2} = \frac{a}{b}

where aa and bb are integers, b0b \neq 0, and gcd(a,b)=1\gcd(a, b) = 1.

Step 2: Square Both Sides

Square both sides of the equation to eliminate the square root:

2=a2b22 = \frac{a^2}{b^2}

Step 3: Multiply by b2b^2

To remove the fraction, multiply both sides by b2b^2:

2b2=a22b^2 = a^2

This equation implies that a2a^2 is an even number because it is equal to 2b22b^2, which is clearly even (since it's two times an integer).

Step 4: Analyze the Parity of aa

If a2a^2 is even, then aa itself must also be even (since the square of an odd number is odd). Therefore, we can write aa as a=2ka = 2k for some integer kk.

Step 5: Substitute a=2ka = 2k into the Equation

Substitute a=2ka = 2k into the equation 2b2=a22b^2 = a^2:

2b2=(2k)22b^2 = (2k)^2

2b2=4k22b^2 = 4k^2

Step 6: Simplify the Equation

Divide both sides of the equation by 2:

b2=2k2b^2 = 2k^2

This equation implies that b2b^2 is also even, so bb must also be even.

Step 7: Reach a Contradiction

Now we have found that both aa and bb are even. This means that aa and bb have at least 2 as a common factor. But this contradicts our original assumption that ab\frac{a}{b} is in the simplest form (i.e., gcd(a,b)=1\gcd(a, b) = 1).

Conclusion

The contradiction arises from our initial assumption that 2\sqrt{2} is rational. Therefore, we must conclude that 2\sqrt{2} is not a rational number.

Would you like further details or have any questions?

Related Questions:

  1. Can you prove that 3\sqrt{3} is not rational?
  2. What is the difference between rational and irrational numbers?
  3. How do we prove that π\pi is irrational?
  4. Can the square root of any non-perfect square be a rational number?
  5. How does the proof for 2\sqrt{2} being irrational differ from proving 4\sqrt{4} is rational?

Tip:

When proving by contradiction, always ensure that your assumption contradicts a known fact or leads to an impossible situation. This method is powerful for proving irrationality and other mathematical properties.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Number Theory
Proof by Contradiction
Irrational Numbers

Formulas

-

Theorems

-

Suitable Grade Level

Grades 10-12