Approximate Location On Number Line

[mathdad]

Show the approximate location of the irrational number
[(3 - sqrt{5})/2]^(1/2) on a number line without a calculator. Then use a calculator to confirm your approximation.

I can find the approximate location using a calculator. How is this done without a calculator?

 

[MarkFL]

We could start by writing:

\(\displaystyle x=\sqrt{\frac{5-\sqrt{3}}{2}}\)

We know we will only be interested in positive values.

Square:

\(\displaystyle x^2=\frac{5-\sqrt{3}}{2}\)

\(\displaystyle 2x^2=5-\sqrt{3}\)

\(\displaystyle 5-2x^2=\sqrt{3}\)

This tells us we are interested in the smaller of any positive roots, when we square again:

\(\displaystyle 25-20x^2+4x^4=3\)

Standard form:

\(\displaystyle f(x)=2x^4-10x^2+11=0\)

We find:

\(\displaystyle f(1)=3\)

\(\displaystyle f(2)=3\)

We should suspect that there are two roots between 1 and 2, and let's cut in between and see if we get a negative value...

\(\displaystyle f\left(\frac{3}{2}\right)=-11/8\)

Yes, it's negative, so the root we want must be in between.

So, we know:

\(\displaystyle 1<x<\frac{3}{2}\)

\(\displaystyle f\left(\frac{5}{4}\right)=33/128\)

Now we know:

\(\displaystyle \frac{5}{4}<x<\frac{3}{2}\)

\(\displaystyle f\left(\frac{13}{10}\right)=-\frac{939}{5000}\)

Thus:

\(\displaystyle \frac{5}{4}<x<\frac{13}{10}\)

We could continue to get whatever desired accuracy we want, but at this point we know $1.25<x<1.3$, which is a reasonable approximation. :D

 

[Greg]

Note that $\sqrt{\frac{3-\sqrt5}{2}}=\left|\frac{1-\sqrt5}{2}\right|$

 

[MarkFL]

Note that $\sqrt{\frac{3-\sqrt5}{2}}=\left|\frac{1-\sqrt5}{2}\right|$

Yeah, I messed up the value we want by interchanging the 3 and the 5...I hate when that happens. :D

That would simply things greatly. (Yes)

\(\displaystyle x=\sqrt{\frac{3-\sqrt5}{2}}=\sqrt{\frac{6-2\sqrt5}{4}}=\frac{\sqrt{1-2\sqrt{5}+5}}{2}=\frac{\sqrt{(\sqrt{5}-1)^2}}{2}=\frac{\sqrt{5}-1}{2}\)

\(\displaystyle 2<\sqrt{5}<3\)

\(\displaystyle 1<\sqrt{5}-1<2\)

\(\displaystyle \frac{1}{2}<\frac{\sqrt{5}-1}{2}<1\)

Begin with a narrower interval containing $\sqrt{5}$, and you'll end up with a better approximation for the requested value. :D

 

[mathdad]

We could start by writing:

\(\displaystyle x=\sqrt{\frac{5-\sqrt{3}}{2}}\)

We know we will only be interested in positive values.

Square:

\(\displaystyle x^2=\frac{5-\sqrt{3}}{2}\)

\(\displaystyle 2x^2=5-\sqrt{3}\)

\(\displaystyle 5-2x^2=\sqrt{3}\)

This tells us we are interested in the smaller of any positive roots, when we square again:

\(\displaystyle 25-20x^2+4x^4=3\)

Standard form:

\(\displaystyle f(x)=2x^4-10x^2+11=0\)

We find:

\(\displaystyle f(1)=3\)

\(\displaystyle f(2)=3\)

We should suspect that there are two roots between 1 and 2, and let's cut in between and see if we get a negative value...

\(\displaystyle f\left(\frac{3}{2}\right)=-11/8\)

Yes, it's negative, so the root we want must be in between.

So, we know:

\(\displaystyle 1<x<\frac{3}{2}\)

\(\displaystyle f\left(\frac{5}{4}\right)=33/128\)

Now we know:

\(\displaystyle \frac{5}{4}<x<\frac{3}{2}\)

\(\displaystyle f\left(\frac{13}{10}\right)=-\frac{939}{5000}\)

Thus:

\(\displaystyle \frac{5}{4}<x<\frac{13}{10}\)

We could continue to get whatever desired accuracy we want, but at this point we know $1.25<x<1.3$, which is a reasonable approximation. :D

What a great reply! You are great!

- - - Updated - - -

Yeah, I messed up the value we want by interchanging the 3 and the 5...I hate when that happens. :D

That would simply things greatly. (Yes)

\(\displaystyle x=\sqrt{\frac{3-\sqrt5}{2}}=\sqrt{\frac{6-2\sqrt5}{4}}=\frac{\sqrt{1-2\sqrt{5}+5}}{2}=\frac{\sqrt{(\sqrt{5}-1)^2}}{2}=\frac{\sqrt{5}-1}{2}\)

\(\displaystyle 2<\sqrt{5}<3\)

\(\displaystyle 1<\sqrt{5}-1<2\)

\(\displaystyle \frac{1}{2}<\frac{\sqrt{5}-1}{2}<1\)

Begin with a narrower interval containing $\sqrt{5}$, and you'll end up with a better approximation for the requested value. :D

Another great reply. You are awesome!