Simplifying a square root in a fraction, part of midpoint formula

[OMGMathPLS]

I have:

sq rt 2 +sq rt 2 over 2 , sq rt 5 + sq rt 5 over 2

I got (sq rt 4 over 2, and 0) = 1, 0

but the answer is actually (sq rt 2, 0)

so is my answer still wrong?

 

[MarkFL]

First, you cannot state:

\(\displaystyle \sqrt{a}+\sqrt{b}=\sqrt{a+b}\)

This is not true in general.

But, this kind of mistake is so common, it is referred to as the "Freshman's Dream."

What you want to do is:

\(\displaystyle \frac{\sqrt{2}+\sqrt{2}}{2}=\frac{2\sqrt{2}}{2}=\sqrt{2}\)

Think about it...you are finding the arithmetic mean, or average, of two numbers that happen to have the same value...wouldn't you then expect the average to be the same as the two identical values?

 

[OMGMathPLS]

no, it doesn't really make sense because you guys said that it can be broken apart... so it can't be combined?

You said that sq rt 12 can become sq rt 3 + sq rt 4

something similar to that. That's probably wrong, but I'm referring to an earlier question i asked. So it just seems natural you can combine them. sorry I can't see through beyond dream but thanks for explaining it.

 

[MarkFL]

no, it doesn't really make sense because you guys said that it can be broken apart... so it can't be combined?

You said that sq rt 12 can become sq rt 3 + sq rt 4

something similar to that. That's probably wrong, but I'm referring to an earlier question i asked. So it just seems natural you can combine them. sorry I can't see through beyond dream but thanks for explaining it.

What you can do is:

\(\displaystyle \sqrt{12}=\sqrt{4\cdot3}=\sqrt{4}\cdot\sqrt{3}=2\sqrt{3}\)

But, you cannot do this with addition. This has to do with rules of exponents (since a square root is a rational exponent).

It is true that:

\(\displaystyle (ab)^c=a^cb^c\)

But it is not true that:

\(\displaystyle (a+b)^c=a^c+b^c\)

This is the "Freshman's Dream" in its most common form. :D

 

[OMGMathPLS]

Ok that makes sense. I can only be done with mult and division but not sub and add.

Because a term is only addition but a factor is like multiplication. And more can be done with that.

 

[MarkFL]

I wanted to reiterate my point earlier about the mean of two identical values being the same as the two values. Suppose there are two people and both weigh 150 lb. Wouldn't you then expect the average weight of the two people to be 150 lb.?

Suppose our two data values are $x$. Then the mean is:

\(\displaystyle \overline{x}=\frac{x+x}{2}=\frac{2x}{2}=x\)

So, this corresponds to the fact that if we are given two points in the $xy$-plane and both points are on the same horizontal or vertical line, then the mid-point will also be on this line. Does this make sense intuitively?

 

[OMGMathPLS]

You make sense. I am just new at this.

I don't know why the sq rt had to stay.

 

[MarkFL]

You make sense. I am just new at this.

I don't know why the sq rt had to stay.

Both points lie along the line $x=\sqrt{2}$, and so the $x$-coordinate of the mid-point must lie along this line also, and so the $x$-coordinate of the mid-point has to be $\sqrt{2}$. :D

 

[OMGMathPLS]

Both points lie along the line $x=\sqrt{2}$, and so the $x$-coordinate of the mid-point must lie along this line also, and so the $x$-coordinate of the mid-point has to be $\sqrt{2}$. :D

because it's already in that form
keep it that way