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發問:
Suppose q is a rational number such that 1 < q ≤ 3-sqrt(2). Write q=m/n, where m,n ∈ N and gcd(m,n)=1.Prove that sqrt(2) ia a positive distance from q. Specifically, show that |sqrt(2)-m/n| ≥ 1/(3n^2) (**)Hint: First prove that if gcd(m,n)=1 then 2n^2 and m^2 are distinct integers, then "rationalize the... 顯示更多 Suppose q is a rational number such that 1 < q ≤ 3-sqrt(2). Write q=m/n, where m,n ∈ N and gcd(m,n)=1. Prove that sqrt(2) ia a positive distance from q. Specifically, show that |sqrt(2)-m/n| ≥ 1/(3n^2) (**) Hint: First prove that if gcd(m,n)=1 then 2n^2 and m^2 are distinct integers, then "rationalize the numerator" in the LHS of (**) how to prove if gcd(m,n)=1 then 2n^2 and m^2 are distinct integers? I have to assume sqrt(2) is a rational number? how to "rationalize the numerator" |sqrt(2)-m/n|?? Thanks
最佳解答:
claim : If gcd(m,n) = 1 , where m,n ∈ N Then 2n2 and m2 are distinct integers pf : Suppose m2 = 2n2 Then 2︱m So we may assume that m = 2k , for some integer k. m2 = 2n2 implies 4k2 = 2n2 n2 = 2k2 So 2︱n 2︱m and 2︱n implies gcd(m,n) = 2 , a contradiction. Hence, m2 ≠ 2n2 # Since 2n2 and m2 are distinct integers, ︱2n2 - m2︱≧ 1 ..... (1) m/n = q ≦ 3 - √2 √2 + m/n ≦ 3 1/( √2 + m/n ) ≧ 1/3 ..... (2) ︱√2 - m/n︱ =︱n√2 - m︱/ n =︱( n√2 - m )( n√2 + m )︱/ [ n( n√2 + m ) ] , this step called "rationalize the numerator" =︱2n2 - m2︱/ [ n( n√2 + m ) ] ≧ 1 / [ n ( n√2 + m ) ] , by (1) = 1 / [ n2 ( √2 + m/n ) ] = ( 1 / n2 )[ 1 / ( √2 + m/n ) ] ≧ ( 1 / n2 )( 1 / 3 ) , by (2) = 1 / ( 3n2 ) Q.E.D.
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Suppose q is a rational number such that...?發問:
Suppose q is a rational number such that 1 < q ≤ 3-sqrt(2). Write q=m/n, where m,n ∈ N and gcd(m,n)=1.Prove that sqrt(2) ia a positive distance from q. Specifically, show that |sqrt(2)-m/n| ≥ 1/(3n^2) (**)Hint: First prove that if gcd(m,n)=1 then 2n^2 and m^2 are distinct integers, then "rationalize the... 顯示更多 Suppose q is a rational number such that 1 < q ≤ 3-sqrt(2). Write q=m/n, where m,n ∈ N and gcd(m,n)=1. Prove that sqrt(2) ia a positive distance from q. Specifically, show that |sqrt(2)-m/n| ≥ 1/(3n^2) (**) Hint: First prove that if gcd(m,n)=1 then 2n^2 and m^2 are distinct integers, then "rationalize the numerator" in the LHS of (**) how to prove if gcd(m,n)=1 then 2n^2 and m^2 are distinct integers? I have to assume sqrt(2) is a rational number? how to "rationalize the numerator" |sqrt(2)-m/n|?? Thanks
最佳解答:
claim : If gcd(m,n) = 1 , where m,n ∈ N Then 2n2 and m2 are distinct integers pf : Suppose m2 = 2n2 Then 2︱m So we may assume that m = 2k , for some integer k. m2 = 2n2 implies 4k2 = 2n2 n2 = 2k2 So 2︱n 2︱m and 2︱n implies gcd(m,n) = 2 , a contradiction. Hence, m2 ≠ 2n2 # Since 2n2 and m2 are distinct integers, ︱2n2 - m2︱≧ 1 ..... (1) m/n = q ≦ 3 - √2 √2 + m/n ≦ 3 1/( √2 + m/n ) ≧ 1/3 ..... (2) ︱√2 - m/n︱ =︱n√2 - m︱/ n =︱( n√2 - m )( n√2 + m )︱/ [ n( n√2 + m ) ] , this step called "rationalize the numerator" =︱2n2 - m2︱/ [ n( n√2 + m ) ] ≧ 1 / [ n ( n√2 + m ) ] , by (1) = 1 / [ n2 ( √2 + m/n ) ] = ( 1 / n2 )[ 1 / ( √2 + m/n ) ] ≧ ( 1 / n2 )( 1 / 3 ) , by (2) = 1 / ( 3n2 ) Q.E.D.
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