In the previous part we understood clearly the difference between a Rational Number and an IRRATIONAL number.
So, the question is why √2 is not a Rational Number?
Or to ask the question differently can √2 be represented as a/b where a and b are integers and have no common factor?
Let’s assume that √2 is a Rational Number. So we must be able to represent √2 as a/b and a and b have no common factors.
So,
- √2 = a/b
2 = a2/ b2
2b2= a2
a2 is a multiple of 2 (i.e. 2 * b2), and so a2 is an even number. That means a is also an even number.
If a is even, then we can write a as a multiple of 2, say for example 2c.
Then 2b2 = (2c)2
2b2 = 4c2
b2 = 2c2
The above says b2 is a multiple of 2 ( i.e. 2 * c2) and so b2 must be even. That means b is also an even number.
In other words both a and b are even. If both are even then both are divisible by 2 and that means both have a common factor other than 1.
Now, according a Rational Number they cannot have a common factor.
That means we cannot express √2 in the form of a Rational Number.
That means you cannot measure the segment hypotenuse in the triangle below with Rational Numbers.
But does that mean these are not numbers. No, they are Real Numbers. For example, in the picture below you would see that the Real Number √2 corresponds to a point on the line:
You may also click here for more detailed understanding of the value √2.
To put it in lighter terms, the Pythagoreans ran out of numbers when they had to measure such lengths, for their system had only numbers of the form m/n, i.e., the Rational Numbers and they did not know other real numbers (irrational numbers) such as √2.
It is said that when the Pythagoreans realized this shortfall in their system, their rank fell apart and ended up in their downfall.
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Very good information
Good explanation.
Thanks for a brief illustration and requesting good self to share such artciles in this common formum which will benefit the students and parents .
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Mam
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Yokibu Team
To prove that square root of 2 is not a rational number:
Let us assume that √2 were a rational number.
We can say that √2 = a/b where a, b are whole numbers, b not zero, simplified to the lowest terms.
Hence it follows that 2 = a2/b2, or a2 = 2 * b2.
So the square of a is an even number since it is two times a, and therefore a itself is also an even number.
The reason for the above is that if a itself was odd, then a * a would be odd too. Odd number times odd number is always odd.
So, if a itself is an even number, then a is 2 times some other whole number, or a = 2k where k is this other whole number.
If we substitute a = 2k into the original equation 2 = a2/b2, we get this equation:
2 = (2k)2/b2
2 = 4k2/b2
2*b2 = 4k2
b2 = 2k2.
It is inferred from above that b2 is even, from which it follows that b itself is an even number.
Since we started the whole process stating that a/b is simplified to the lowest terms, and since both a and b are even, √2 cannot be a rational number.
gr8