proof that 2 is irrational step by step
Answers
Step-by-step explanation:
Euclid's proof starts with the assumption that √2 is equal to a rational number p/q.
√2=p/q
Squaring both sides,
2=p²/q²
The equation can be rewritten as
2q²=p²
From this equation, we know p² must be even (since it is 2 multiplied by some number). Since p² is an even number, it can be inferred that p is also an even number.
Since p is even, it can be written as 2m where m is some other whole number. This is because the definition of an even number is it can be written as 2 multiplied by a whole number. Substituting p=2m in the above equation:
2q²=p²=(2m)²=4m²
or
2q²=4m²
Dividing both sides of the equation by 2:
q²=2m²
By the same reasoning as before, q² is an even number (since it is written as 2 multiplied by some number). So q is an even number. Let q=2n where n is some whole number. We had assumed √2 to be equal to p/q. So:
√2=p/q=2m/2n
By canceling 2 in the numerator and the denominator of the Right hand side,
√2=m/n
We now have a fraction m/n simpler than p/q.
However, we now find ourselves in a position whereby we can repeat exactly the same process on m/n, and at the end of it, we can generate a simpler function, say g/h. This fraction can be put through the same process again, and the new fraction, say, e/f will be simpler again.
But we know that rational number cannot be simplified indefinitely. There must always be a simplest rational number and the original assumption that √2 is equal to p/q does not obey this rule.
So it can be stated that a contradiction has been reached.
If √2 could be written as a rational number, the consequence would be absurd. So it is true to say that √2 cannot be written in the form p/q. Hence √2 is not a rational number.
Thus, Euclid succeeded in proving that √2 is an Irrational number.
Answer:
Root2 is an irrational number.
Step-by-step explanation:
Here, As per our given question,
Let us assume that root2 is a rational number with its simplest form=a/b and are two integers a and b which are co-primes.
Now,=root2=a/b
On squaring both sides, we get
=2=a^2/b^2
=2b^2=a^2 -(1st)eq.
Here, As 2 is a prime no. and 2 divides b^2,
So, =2 divides a^2
=2 divides a
Now, Let a=2c for integer c,
By putting value of 2 in (1st)eq., we get,
=2b^2=(2c)^2
=2b^2=4c^2
=b^2=2c^2 (As, 2 is on both sides)
Here also,2 is a prime no. and 2 divides 2c^2,
So, =2 divides b^2
=2 divides b
Thus, as we get that a and b both have a common factor which is 2, But as we assumed that a and b are co-primes(numbers which have HCF=1, or only one factor).
This contradiction has arisen due to our wrong assumption that root 2 is rational number.
Hence, root2 is an irrational number.
Thank you.