Prove that there is no rational number q such q^2 = q*q = 2.?

Answer 1

We must prove this by contradiction.

Assume that there DOES exist a rational number q such that

q^2 = 2

Since q is rational, q can be expressed as a quotient of integers; that is,

q = a/b, where a and b are both integers and where b is non-zero.

Let us also assume that a and b have no common factors (because if they did have a common factor, we can reduce the fraction further such that they have no common factors).

Therefore, since
q^2 = 2, then

(a/b)^2 = 2

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

Multiply both sides by b^2, to get

a^2 = 2b^2

This shows that a^2 is a multiple of 2, which means a is a multiple of 2. If a is a multiple of 2, then a can be expressed in the form 2k; therefore,

[2k]^2 = 2b^2
4k^2 = 2b^2. Divide by 2,
2k^2 = b^2

Showing b^2 is a multiple of 2. If b^2 is a multiple of 2, then b has to be a multiple of 2.

Therefore, a is a multiple of 2 and b is a multiple of 2. This is a contradiction, because, as we've stated in the beginning, a and b have no common factors.

Therefore, there exists no rational number q such that q^2 = 2

Answer 2

If q be a rational number such that q^2 = 2
Then q = sqrt.2

So we need to prove by contradiction that sqrt2 is irrational.

If possible, let sqrt. 2 be rational.
Then it can be expressed in the form a/b where a and b are integers and have no common factors other than 1 and b is not 0.

sqrt. 2 = a / b
2 = a^2 / b^2
a^2 = 2 * b^2
a^2 is even. So a is even.
Let a = 2k where k is an integer.
2 = (2k)^2 / b^2
2 = 4k^2 / b^2
b^2 = 2 * k^2
b^2 is even. So b is even.

This is a contradiction as a and b have no common factors other than 1.

Thus our assumption is wrong.

Hence sqrt. 2 is irrtaional.

Hope this helps.

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Answer 3

This is an extrememly fancy way of asking us to prove the irrationality of √2.

Notice the question well:
q^2 = 2
q = √2

We must prove that q is not rational. Meaning √2 is irrational.

Assume that q is rational. If our assumption is contradicted in any step in the working, then it means that our assumption is wrong, and q is irrational.

q is rational. Meaning q = a/b

where a, b are integers, b ≠ 0, a and b are coprime.
Coprime means that the HCF (Highest Common Factor) is 1.

a/b = √2
a^2/b^2 = 2
a^2 = 2b^2 ..... (1)

There is a law which states that is the square of any integer is even, then the number must be even. 2b^2 is necessarily even, since it is an integer multiplied by 2.
And since a^2 = an even number, a is even.

So, we can write a as 2c, where c is another integer.
a = 2c

(1) becomes,

(2c)^2 = 2b^2
4c^2 = 2b^2
b^2 = 2c^2

b is also even.

Now a and b are even. Meaning they have 2 as a common factor. But this contradicts our earlier assumption that a and b are coprime (HCF = 1). Since there is no mistanke in the working, what we have assumed must be wrong.
So, q is irrational.

Answer 4

q is square root of 2

Answer 5

1.414213562 ^2 = 2 What do you think?