Examine whether root 2 rational or irrationalExamine whether root 2 rational or irrational
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Answer:
Given √2
Given √2To prove: √2 is an irrational number.
Given √2To prove: √2 is an irrational number.Proof:
Given √2To prove: √2 is an irrational number.Proof:Let us assume that √2 is a rational number.
Given √2To prove: √2 is an irrational number.Proof: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
Given √2To prove: √2 is an irrational number.Proof: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
Given √2To prove: √2 is an irrational number.Proof: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
Given √2To prove: √2 is an irrational number.Proof: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
Given √2To prove: √2 is an irrational number.Proof: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
Given √2To prove: √2 is an irrational number.Proof: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,
Given √2To prove: √2 is an irrational number.Proof: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
Given √2To prove: √2 is an irrational number.Proof: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 = p…..(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 = 4m……..(2)
……..(2)From equations (1) and (2), we get,
……..(2)From equations (1) and (2), we get,2q² = 4m²
……..(2)From equations (1) and (2), we get,2q² = 4m²⇒ q² = 2m²
……..(2)From equations (1) and (2), we get,2q² = 4m²⇒ q² = 2m²⇒ q² is a multiple of 2
……..(2)From equations (1) and (2), we get,2q² = 4m²⇒ q² = 2m²⇒ q² is a multiple of 2⇒ q is a multiple of 2
……..(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)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.
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Answer:
Given √2
Given √2To prove: √2 is an irrational number.
Given √2To prove: √2 is an irrational number.Proof:
Given √2To prove: √2 is an irrational number.Proof:Let us assume that √2 is a rational number.
Given √2To prove: √2 is an irrational number.Proof: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
Given √2To prove: √2 is an irrational number.Proof: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
Given √2To prove: √2 is an irrational number.Proof: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
Given √2To prove: √2 is an irrational number.Proof: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
Given √2To prove: √2 is an irrational number.Proof: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
Given √2To prove: √2 is an irrational number.Proof: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,
Given √2To prove: √2 is an irrational number.Proof: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
Given √2To prove: √2 is an irrational number.Proof: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 = p…..(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 = 4m……..(2)
……..(2)From equations (1) and (2), we get,
……..(2)From equations (1) and (2), we get,2q² = 4m²
……..(2)From equations (1) and (2), we get,2q² = 4m²⇒ q² = 2m²
……..(2)From equations (1) and (2), we get,2q² = 4m²⇒ q² = 2m²⇒ q² is a multiple of 2
……..(2)From equations (1) and (2), we get,2q² = 4m²⇒ q² = 2m²⇒ q² is a multiple of 2⇒ q is a multiple of 2
……..(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)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.