Is there a method to prove that the square root of 2 is a rational number not by using a calculator?

Answer 1

Calculators can never prove numbers are irrational, because they're only approximations, and irrational numbers have an infinite number of digits after the decimal.

However, there is a way to prove that sqrt(2) is irrational. This is known as a proof by contradiction.

What we do is assume that sqrt(2) is rational, and prove that this cannot happen.

Assume sqrt(2) is rational. Then,

sqrt(2) can be expressed as (a/b), for integers a and b and for b not equal to 0. After all, this is the definition of a rational number.

Let us also assume that a and b have no common factors. If a and b had a common factor, then we can further reduce it such that they don't have a common factor (i.e. 4/6 can be reduced to 2/3, and they no longer have a common factor).

sqrt(2) = a/b

Square both sides,

2 = (a^2)/(b^2)

Now, multiply both sides by b^2.

2b^2 = a^2

This tells us that a^2 is an even number, because a^2 is expressed as 2 times something. Therefore, a^2 is even. Since a^2 is even, it follows that a is even.

Now that we know a is even, a can be expressed as 2k, for some integer constant k. That is,

a = 2k.

Therefore, since we have 2b^2 = a^2, then, substituting a = 2k, we have

2b^2 = [2k]^2. Therefore,
2b^2 = 4k^2. Dividing both sides by 2, we get
b^2 = 2k^2.

This tells us b^2 is even. Because b^2 is even, b must be even too.

Therefore, a is even and b is even. But we just stated at the beginning of our proof that "a" and b do not have any common factors, and we've just shown that "a" and b have the common factor of 2. This is a contradiction.

Therefore, sqrt(2) is irrational.

Answer 2

You can't prove that the square root of 2 is a rational number, because it isn't. It is an irrational number.

A rational number is one which can be written as a ratio: something divided by something, where the two somethings are both whole numbers.

For example, the square root of 2 is approximately 297 divided by 211. Is it possible to find two such numbers which give you the square root exactly?

Let's assume it is possible. There may be a number of different pairs that give the same result, so let's pick the pair which have has the smallest number below the line. Call them a and b.

We have:
sqrt(2) = a/b
Squaring this we get:
2 = (a/b)² = a² / b²
Multiplying by b² we get:
2a² = b²

So b² is an even number. If b² is even, then b is also even (even*even = even, odd*odd = odd). If b is even, we can divide it by 2. Call this c:
c = b/2 or b = 2c

So now we have:
2a² = b² = (2c)² = 4c²
a² = 2c²
a²/c² = 2
a/c = sqrt(2)

So we have another pair, a and c which also give us the square root of 2. c is smaller than b and we picked b to be the smallest possible.

So if b is the smallest, then there is an even smaller number, c. THis doesn't make any sense, so our original statement, that there can be any pair of numbers must be wrong.

So there is no pair of numbers which when expressed as a ratio gives the square root of 2. It is irrational.

Answer 3

V2 is irrational, your calculator isn't much help proving this.
Consider the following proof,

let V2 = p/q
if you recall a rational number occurs in the form p/q , with p,q integers and q non negative or zero.
then 2 = p^2 / q^2
=> 2q^2 = p^2
p^2 is an even number(in the integers) owing to the 2 on the left hand side. But for a square of an integer to be even, the original integer has to have a factor of 2, which when squared gives a factor of 4 to p^2.
This implies that q^2 is even. But similarly q^2 must have a factor of 4. Meaning p and q have a common factor of 2. Then p/q cannot be in its lowest terms. Therefore V2 cannot be put in the form p/q making it irrational.

Hope this helps!

Answer 4

square root of 2 is irrational number.

to prove it, imagine it isn't, so
sqrt(2) = a/b
where a and b are integers, and fraction cannot be simplified any further (i.e. a and b have no shared prime factors)

square the equation above, you get:
2=a^2/b^2

now let's go back to prime factors - a^2 has them same as a just two of each, same for b^2. But we know that a and b have no common factors, so we cannot simplify a^2/b^2 to an integer.
Contradition.

Answer 5

i took square roots in high school back in the sixties. we didn't have calculators so we had to use pencil and paper to do any square root and the proofs. you should try it. of course it'll take yoiu days to get to the square root of 2, but proving it is easy ___NOT!!