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√2 is irrational

Prove that √2 is irrational number i.e. it can not be represented as a fraction a÷b where a and b are integers, b>0.

This is a very clever proof as a practice on proof by contradiction.

If you believe that √2 is a rational number, then I’ll ask you to come with this fraction a÷b but first reduce it to lowest terms. And I’ll convince you at the end that a and b must be both even. Does this mean you are a liar? 😉 – No it just means that there is a contradiction and thus √2 can not be rational (and that a÷b you have suggested is a fake! 😉 ).

So let’s assume that √2 can be represented as a÷b (You have just claimed that) where a and b are in their lowest terms. This means that a and b can not both be even numbers, because this means they are not yet in their lowest terms as they can be divided by 2.

√2 = a÷b (square both sides)

2 = a²÷b²

2b² = a²

The above equation implies that a² is an even number. Which subsequently means that a is also even.

Because a is even, we can assume that  a = 2k

2b² = (2k)²

2b² = 4k²

b² = 2k²

Again the above equation implies that b² is even and thus b is also even.

But being a and b even numbers is a contradiction to the assumption at the beginning that a and b are in their lowest terms.

What does it mean that an assumption contradicts itself? – Certainly mean that this assumption (√2 is rational) is wrong.

So √2 is irrational number.


The proof is done at this point, But I’d like to mention some historical notes:

1- It was one of the most surprising discoveries of the Pythagorean School of Greek mathematicians that there are  irrational numbers.  According to Courant and Robbins in  “What is Mathematics”:   This revelation was a scientific event of the highest importance. Quite possibly it marked the origin of what we consider the specifically Greek contribution to rigorous procedure in mathematics.

2- The Egyptian mathematician Abū Kāmil (c. 850–930) was the first to accept irrational numbers as solutions to quadratic equations or as coefficients in an equation, often in the form of square roots, cube roots and fourth roots.

Question

Can you prove with a similar technique that the square root of 3 is irrational?

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  1. Sherif Kamal
    February 16, 2009 at 12:06 AM

    Can you prove with a similar technique that the square root of 3 is irrational?

    i’ll try 🙂

    So let’s assume that √3 can be represented as a÷b (You have just claimed that) ,
    where a and b are in their lowest terms. This means that a and b can not both be multiple of 3 numbers, because this means they are not yet in their lowest terms as they can be divided by 3.

    √3 = a÷b (square both sides)

    3 = a²÷b²

    3b² = a²

    The above equation implies that a² is a multiple of 3 number.

    So, we can assume that a = 3k

    2b² = (3k)²

    2b² = 9k²

    b² = 3k²

    Again the above equation implies that b² is even and thus b is also a multiple of 3 number.

    But being a and b a multiple of 3 numbers is a contradiction to the assumption at the beginning that a and b are in their lowest terms.

    So √3 is irrational number.

  2. Ahmed Abdullah
    February 16, 2009 at 1:06 AM

    Nice try Sherif!

    But you missed in reducing one of the equations

    So, we can assume that a = 3k
    2b² = (3k)²

    You probably meant 3b² = (3k)²

  3. Sherif Kamal
    February 16, 2009 at 1:11 AM

    So Sorry 😦 :
    i missed my concentration during coping the paragraph :

    3b² = (3k)²

    3b² = 9k²

    b² = 3k²

  4. Ahmed Abdullah
    February 16, 2009 at 1:31 AM

    OK. I agree with you until your last equation

    b² = 3k²

    You then claim that:

    Again the above equation implies that b² is even.

    That is not true. If b² = 3k² then b² is odd and thus b is odd.

    Rethink about it.

  5. Sherif Kamal
    February 16, 2009 at 1:35 AM

    The same error :((

    Again the above equation implies that b² is multiple of 3 and thus b is also a multiple of 3 number.

  6. Ahmed Abdullah
    February 16, 2009 at 1:40 AM

    Again the above equation implies that b² is multiple of 3 and thus b is also a multiple of 3 number.

    OK. But you are using a wrong reasoning in this sentence.
    If b² is a multiple of 3 then b is NOT necessarily a multiple of 3.

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