find√23 with full solution
Answers
Answered by
2
Answer:
√23≈1151240=4.7958¯3
Explanation:
23 is a prime number, so it is not possible to simplify its square root, which is an irrational number a little less than 5=√25
As such it is not expressible in the form pq for integers p,q.
We can find rational approximations as follows:
23=52−2
is in the form n2−2
The square root of a number of the form n2−2can be expressed as a continued fraction of standard form:
√n2−2=[(n−1);¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯1,(n−2),1,(2n−2)]
In our example n=5 and we find:
√23=[4;¯¯¯¯¯¯¯¯¯¯¯¯¯¯1,3,1,8]=4+11+13+11+18+11+13+11+...
To use this to derive a good approximation for √23 terminate it early, just before one of the 8's. For example:
√23≈[4;1,3,1,8,1,3,1]=4+11+13+11+18+11+13+11=1151240=4.7958¯3
With a calculator, we find:
√23≈4.79583152
√23≈1151240=4.7958¯3
Explanation:
23 is a prime number, so it is not possible to simplify its square root, which is an irrational number a little less than 5=√25
As such it is not expressible in the form pq for integers p,q.
We can find rational approximations as follows:
23=52−2
is in the form n2−2
The square root of a number of the form n2−2can be expressed as a continued fraction of standard form:
√n2−2=[(n−1);¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯1,(n−2),1,(2n−2)]
In our example n=5 and we find:
√23=[4;¯¯¯¯¯¯¯¯¯¯¯¯¯¯1,3,1,8]=4+11+13+11+18+11+13+11+...
To use this to derive a good approximation for √23 terminate it early, just before one of the 8's. For example:
√23≈[4;1,3,1,8,1,3,1]=4+11+13+11+18+11+13+11=1151240=4.7958¯3
With a calculator, we find:
√23≈4.79583152
Similar questions