prove that √2 is irrational.
Answers
Let us assume that √2 is a rational number.
Let us assume that √2 is a rational number.So it can be expressed in the form p/q where p, q are co-prime integers and q≠0
Let us assume that √2 is a rational number.So it can be expressed in the form p/q where p, q are co-prime integers and q≠0√2 = p/q
Let us assume that √2 is a rational number.So it can be expressed in the form p/q where p, q are co-prime integers and q≠0√2 = p/qHere p and q are coprime numbers and q ≠ 0
Let us assume that √2 is a rational number.So it can be expressed in the form p/q where p, q are co-prime integers and q≠0√2 = p/qHere p and q are coprime numbers and q ≠ 0Solving
Let us assume that √2 is a rational number.So it can be expressed in the form p/q where p, q are co-prime integers and q≠0√2 = p/qHere p and q are coprime numbers and q ≠ 0Solving√2 = p/q
Let us assume that √2 is a rational number.So it can be expressed in the form p/q where p, q are co-prime integers and q≠0√2 = p/qHere p and q are coprime numbers and q ≠ 0Solving√2 = p/qOn squaring both the side we get,
Let us assume that √2 is a rational number.So it can be expressed in the form p/q where p, q are co-prime integers and q≠0√2 = p/qHere p and q are coprime numbers and q ≠ 0Solving√2 = p/qOn squaring both the side we get,=>2 = (p/q)²
=> 2q² = p²……………………………..(1)
……………………………..(1)p2/2 = q2
……………………………..(1)p2/2 = q2So 2 divides p and p is a multiple of 2.
……………………………..(1)p2/2 = q2So 2 divides p and p is a multiple of 2.⇒ p = 2m
……………………………..(1)p2/2 = q2So 2 divides p and p is a multiple of 2.⇒ p = 2m⇒ p² = 4m² ………………………………..(2)
……………………………..(1)p2/2 = q2So 2 divides p and p is a multiple of 2.⇒ p = 2m⇒ p² = 4m² ………………………………..(2)From equations (1) and (2), we get,
……………………………..(1)p2/2 = q2So 2 divides p and p is a multiple of 2.⇒ p = 2m⇒ p² = 4m² ………………………………..(2)From equations (1) and (2), we get,2q² = 4m²
……………………………..(1)p2/2 = q2So 2 divides p and p is a multiple of 2.⇒ p = 2m⇒ p² = 4m² ………………………………..(2)From equations (1) and (2), we get,2q² = 4m²⇒ q² = 2m²
……………………………..(1)p2/2 = q2So 2 divides p and p is a multiple of 2.⇒ p = 2m⇒ p² = 4m² ………………………………..(2)From equations (1) and (2), we get,2q² = 4m²⇒ q² = 2m²⇒ q² is a multiple of 2
……………………………..(1)p2/2 = q2So 2 divides p and p is a multiple of 2.⇒ p = 2m⇒ p² = 4m² ………………………………..(2)From equations (1) and (2), we get,2q² = 4m²⇒ q² = 2m²⇒ q² is a multiple of 2⇒ q is a multiple of 2
……………………………..(1)p2/2 = q2So 2 divides p and p is a multiple of 2.⇒ p = 2m⇒ p² = 4m² ………………………………..(2)From equations (1) and (2), we get,2q² = 4m²⇒ q² = 2m²⇒ q² is a multiple of 2⇒ q is a multiple of 2Hence, p, q have a common factor 2. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number
……………………………..(1)p2/2 = q2So 2 divides p and p is a multiple of 2.⇒ p = 2m⇒ p² = 4m² ………………………………..(2)From equations (1) and (2), we get,2q² = 4m²⇒ q² = 2m²⇒ q² is a multiple of 2⇒ q is a multiple of 2Hence, p, q have a common factor 2. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number√2 is an irrational number.
Hello... lets do it cmon...
Let's suppose √2 is a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero.
We additionally assume that this a/b is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in simplest terms, both of a and b cannot be even. One or both must be odd. Otherwise, we could simplify a/b further.
From the equality √2 = a/b it follows that 2 = a2/b2, or a2 = 2 · b2. So the square of a is an even number since it is two times something.
From this we know that a itself is also an even number. Why? Because it can't be odd; if a itself was odd, then a · a would be odd too. Odd number times odd number is always odd. Check it if you don't believe me!
Okay, if a itself is an even number, then a is 2 times some other whole number. In symbols, a = 2k where k is this other number. We don't need to know what k is; it won't matter. Soon comes the contradiction.
If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:
2 = (2k)2/b2
2 = 4k2/b2
2*b2 = 4k2
b2 = 2k2
This means that b2 is even, from which follows again that b itself is even. And that is a contradiction!!!
WHY is that a contradiction? Because we started the whole process assuming that a/b was simplified to lowest terms, and now it turns out that a and b both would be even. We ended at a contradiction; thus our original assumption (that √2 is rational) is not correct. Therefore √2 cannot be rational.
have a grt day ahead buddy...