surd explanation thing i dont understand

[red and white kop!]

show that sqrt(N+1) - sqrt(N) = 1/[sqrt(N+1) + sqrt(N)].
Use this to explain why sqrt(101) is close to, but slightly less than, 10.05.

the first part of this question was kinda tricky for me cos I just didn't see the point of it; i ended up backtracking from the RHS to get the LHS and turned that proof around. maybe this was wrong because it gave me an easy way out but i might have missed the point cos i am totally lost in the second part of the problem. i turned up with something sketchy like sqrt(100 +1) - sqrt(100) = (i don't know what) but this was just in an attempt to match the initial form of the expression.
any help?

 

[Deleted member 4993]

show that sqrt(N+1) - sqrt(N) = 1/[sqrt(N+1) + sqrt(N)].
Use this to explain why sqrt(101) is close to, but slightly less than, 10.05.

the first part of this question was kinda tricky for me cos I just didn't see the point of it; i ended up backtracking from the RHS to get the LHS and turned that proof around. maybe this was wrong because it gave me an easy way out but i might have missed the point cos i am totally lost in the second part of the problem. i turned up with something sketchy like sqrt(100 +1) - sqrt(100) = (i don't know what) but this was just in an attempt to match the initial form of the expression.
any help?

\(\displaystyle \frac{1}{\sqrt {N+1} \ \ + \ \ \sqrt{N}} \ \ < \ \ \frac{1}{2\cdot \sqrt{N}}\)

In your case N = 100

 

[soroban]

Hello, red and white kop!!

\(\displaystyle \text{(a) Show that: }\;\sqrt{N+1} - \sqrt{N} \:=\: \frac{1}{\sqrt{N+1} + \sqrt{N}}\)

(b) Use this to explain why \(\displaystyle \sqrt{101}\) is close to, but slightly less than, \(\displaystyle 10.05\)

\(\displaystyle \text{(a) Multiply the left side by }\,\frac{\sqrt{N+1} + \sqrt{N}}{\sqrt{N+1} + \sqrt{N}}\)

. . \(\displaystyle \frac{\sqrt{N+1} - \sqrt{N}}{1}\cdot\frac{\sqrt{N+1} + \sqrt{N}}{\sqrt{N+1} + \sqrt{N}} \;=\; \frac{(N+1) - N}{\sqrt{N+1} + \sqrt{N}} \;=\;\frac{1}{\sqrt{N+1} + \sqrt{N}}\)



\(\displaystyle \text{(b) We have: }\;\sqrt{N+1} \;=\;\sqrt{N} + \frac{1}{\sqrt{N+1} + \sqrt{N}}\)


\(\displaystyle \text{Let }N = 100\!:\;\;\sqrt{101} \;=\;\sqrt{100} + \frac{1}{\sqrt{101} + \sqrt{100}}\)

\(\displaystyle \text{Then we have: }\;\sqrt{101} \;=\;10 + \frac{1}{\sqrt{101} + 10}\;\;\;[1]\)


. . \(\displaystyle \text{Note that: }\:101 \:>\:100\)

. . \(\displaystyle \text{Then: }\:\sqrt{101} \;\;^>_{\approx}\;\;\sqrt{100} \quad\Rightarrow\quad \sqrt{101} \;\;^>_{\approx}\;\;10\)

. . \(\displaystyle \text{Add 10: }\;\sqrt{101} + 10 \;\;^>_{\approx}\;\;20\)

. . \(\displaystyle \text{Take reciprocals: }\:\frac{1}{\sqrt{101}+10} \;\;^<_{\approx}\;\; \frac{1}{20}\)

. . \(\displaystyle \text{Add 10: }\;10 + \frac{1}{\sqrt{101} + 10} \;\;^<_{\approx}\;\;10 + \frac{1}{20}\)


\(\displaystyle \text{Equate to [1]: }\;\sqrt{101} \;=\;10 + \frac{1}{\sqrt{101} + 10} \;\;^<_{\approx}\;\;10+\frac{1}{20}\)


\(\displaystyle \text{Therefore: }\;\sqrt{101} \;\;^<_{\approx}\;\;10.05\)

 

[Deleted member 4993]

Anotherway (b):

\(\displaystyle \sqrt{N+1} \ \ - \ \ \sqrt{N} \ \ = \ \ \frac{1}{\sqrt{N+1} \ \ + \ \ \sqrt{N}} \ \ < \ \ \frac{1}{2\cdot \sqrt{N}}\)

\(\displaystyle \sqrt{101} \ \ - \ \ \sqrt{100} \ \ < \ \ \frac{1}{2\cdot \sqrt{100}}\)

\(\displaystyle \sqrt{101} \ \ - \ \ 10 \ \ < \ \ 0.05\)

\(\displaystyle \sqrt{101} \ \ < \ \ 10.05\)

 

[red and white kop!]

thanks!