is this an instance of a some arithmoalgeraic (?) rule

[allegansveritatem]

Working out some problems today I came across an equivalence that surprised me. I am sorry I don't have a photo of this and neither do I have access to the tools I need to set it down symbolically. But it is easy to describe. I found that when I had a square root as a numeratorrand a number as a denominator, if I divided the radicand by the denominator and then formed a new fraction with the result of the said division as numerator and the square root as the denominator, those two expressions are equal. For instance the square root of 28 divided by two. ldivide 28 by 2 and you get 14. Put 14 in the numerator of the next fraction and put the square root of 28 in the denominator; then put an = between them and Lo! it turns out to be true! What, if anything is this an instance of. I mean, is there a rule or theorem that covers this? Or is it just that mathematics is so full of these correspondences that it is not worth mentioning.

I have just edited this post. I twisted up my description the first time and untwisted it in the edit.

 

[topsquark]

Working out some problems today I came across an equivalence that surprised me. I am sorry I don't have a photo of this and neither do I have access to the tools I need to set it down symbolically. But it is easy to describe. I found that when I had a square root as a numeratorrand a number as a denominator, if I divided the radicand by the denominator and then formed a new fraction with the result of the said division as numerator and the square root as the denominator, those two expressions are equal. For instance the square root of 28 divided by two. ldivide 28 by 2 and you get 14. Put 14 in the numerator of the next fraction and put the square root of 28 in the denominator; then put an = between them and Lo! it turns out to be true! What, if anything is this an instance of. I mean, is there a rule or theorem that covers this? Or is it just that mathematics is so full of these correspondences that it is not worth mentioning.

\(\displaystyle \dfrac{\sqrt{28}}{2} = \dfrac{14}{\sqrt{28}}\)

This is called "cross multiplication." Given two fractions \(\displaystyle \dfrac{a}{b}\) and \(\displaystyle \dfrac{c}{d}\). They are equal when ad = bc.

Otherwise it can be worked out by manipulating the radicals a bit.

-Dan

 

[Dr.Peterson]

Working out some problems today I came across an equivalence that surprised me. I am sorry I don't have a photo of this and neither do I have access to the tools I need to set it down symbolically. But it is easy to describe. I found that when I had a square root as a numeratorrand a number as a denominator, if I divided the radicand by the denominator and then formed a new fraction with the result of the said division as numerator and the square root as the denominator, those two expressions are equal. For instance the square root of 28 divided by two. ldivide 28 by 2 and you get 14. Put 14 in the numerator of the next fraction and put the square root of 28 in the denominator; then put an = between them and Lo! it turns out to be true! What, if anything is this an instance of. I mean, is there a rule or theorem that covers this? Or is it just that mathematics is so full of these correspondences that it is not worth mentioning.

In general, what you are describing is this: the fraction \(\displaystyle \dfrac{\sqrt{a}}{b}\) is equal to \(\displaystyle \dfrac{\frac{a}{b}}{\sqrt{a}}\). This can be proved by cross multiplication; or obtained by multiplying the numerator and denominator of the original by \(\displaystyle \dfrac{\sqrt{a}}{b}\):

\(\displaystyle \dfrac{\sqrt{a} \cdot \dfrac{\sqrt{a}}{b}}{b \cdot \dfrac{\sqrt{a}}{b}} = \dfrac{\dfrac{\sqrt{a} \cdot \sqrt{a}}{b}}{\sqrt{a}} = \dfrac{\frac{a}{b}}{\sqrt{a}}\)

Of course, this is the opposite of rationalizing the denominator ... and algebra provides many ways to rewrite any expression.

 

[Steven G]

Or is it just that mathematics is so full of these correspondences that it is not worth mentioning.

Whether it is worth mentioning or not is not the point. You discovered it and it therefore can be proven, as Dr Peterson did for you. If you do not understand Dr P's proof, then please tell us where you got stuck.

 

[allegansveritatem]

\(\displaystyle \dfrac{\sqrt{28}}{2} = \dfrac{14}{\sqrt{28}}\)

This is called "cross multiplication." Given two fractions \(\displaystyle \dfrac{a}{b}\) and \(\displaystyle \dfrac{c}{d}\). They are equal when ad = bc.

Otherwise it can be worked out by manipulating the radicals a bit.

-Dan

Thanks for reply. I will have to check this out tomorrow and see how cross multiplication relates to how I did it. I will return.

 

[allegansveritatem]

In general, what you are describing is this: the fraction \(\displaystyle \dfrac{\sqrt{a}}{b}\) is equal to \(\displaystyle \dfrac{\frac{a}{b}}{\sqrt{a}}\). This can be proved by cross multiplication; or obtained by multiplying the numerator and denominator of the original by \(\displaystyle \dfrac{\sqrt{a}}{b}\):

\(\displaystyle \dfrac{\sqrt{a} \cdot \dfrac{\sqrt{a}}{b}}{b \cdot \dfrac{\sqrt{a}}{b}} = \dfrac{\dfrac{\sqrt{a} \cdot \sqrt{a}}{b}}{\sqrt{a}} = \dfrac{\frac{a}{b}}{\sqrt{a}}\)

Of course, this is the opposite of rationalizing the denominator ... and algebra provides many ways to rewrite any expression.

I will have to go through this tomorrow when my brain has reconstituted it self a little. too late now. I recognize my procedure in expression with the fraction in the numerator. I will return when I have gone over all this again tomorrow. thanks very much for reply.

 

[allegansveritatem]

Whether it is worth mentioning or not is not the point. You discovered it and it therefore can be proven, as Dr Peterson did for you. If you do not understand Dr P's proof, then please tell us where you got stuck.

Thanks for reply. I am going to go over Dr Peterson's post and the one above it and will wrap my head around it if it kills me. I see what is being said, I just need to relate it to...I don't know, maybe why it works. I shall return.

 

[JeffM]

Working out some problems today I came across an equivalence that surprised me. I am sorry I don't have a photo of this and neither do I have access to the tools I need to set it down symbolically. But it is easy to describe. I found that when I had a square root as a numeratorrand a number as a denominator, if I divided the radicand by the denominator and then formed a new fraction with the result of the said division as numerator and the square root as the denominator, those two expressions are equal. For instance the square root of 28 divided by two. ldivide 28 by 2 and you get 14. Put 14 in the numerator of the next fraction and put the square root of 28 in the denominator; then put an = between them and Lo! it turns out to be true! What, if anything is this an instance of. I mean, is there a rule or theorem that covers this? Or is it just that mathematics is so full of these correspondences that it is not worth mentioning.

I have just edited this post. I twisted up my description the first time and untwisted it in the edit.

Let's do this in baby steps. It may be boring, but it eliminates all hints of magic.

\(\displaystyle \text {Given } a > 0 \text { and } b \ne 0.\)

\(\displaystyle \dfrac{\sqrt{a}}{b} = \dfrac{\sqrt{a}}{b} * 1.\) No problem there I hope.

\(\displaystyle \dfrac{\sqrt{a}}{b} * 1 = \dfrac{\sqrt{a}}{b} * \dfrac{\sqrt{a}}{\sqrt{a}}.\)

Again, there should be nothing difficult about that.

\(\displaystyle \dfrac{\sqrt{a}}{b} * \dfrac{\sqrt{a}}{\sqrt{a}} = \dfrac{(\sqrt{a})^2}{b\sqrt{a}}.\)

That is just multiplication of fractions.

\(\displaystyle \dfrac{(\sqrt{a})^2}{b\sqrt{a}} = \dfrac{a}{b\sqrt{a}}.\)

Square a square root and you get what?

\(\displaystyle \dfrac{a}{b\sqrt{a}} = \dfrac{a}{b\sqrt{a}} * 1.\)

Any thing equals itself times 1.

\(\displaystyle \dfrac{a}{b\sqrt{a}} * 1 = \dfrac{a}{b\sqrt{a}} * \dfrac{1\ /\ b}{1\ /\ b}.\)

Any non-zero number divided by itself equals 1.

\(\displaystyle \dfrac{a}{b\sqrt{a}}* \dfrac{1\ /\ b}{1\ /\ b} = \dfrac{a\ /\ b}{b\sqrt{a}\ / \ b}.\)

Multiplication of fractions again.

\(\displaystyle \dfrac{a\ /\ b}{b\sqrt{a}\ /\ b} = \dfrac{\dfrac{a}{b}}{\sqrt{a}}.\)

Now if you followed that every step of the way, we just proved that

\(\displaystyle \dfrac{\sqrt{a}}{b} = \dfrac{\dfrac{a}{b}}{\sqrt{a}}.\)

Nothing mysterious, just a lot of basic steps.

 

[allegansveritatem]

Let's do this in baby steps. It may be boring, but it eliminates all hints of magic.

\(\displaystyle \text {Given } a > 0 \text { and } b \ne 0.\)

\(\displaystyle \dfrac{\sqrt{a}}{b} = \dfrac{\sqrt{a}}{b} * 1.\) No problem there I hope.

\(\displaystyle \dfrac{\sqrt{a}}{b} * 1 = \dfrac{\sqrt{a}}{b} * \dfrac{\sqrt{a}}{\sqrt{a}}.\)

Again, there should be nothing difficult about that.

\(\displaystyle \dfrac{\sqrt{a}}{b} * \dfrac{\sqrt{a}}{\sqrt{a}} = \dfrac{(\sqrt{a})^2}{b\sqrt{a}}.\)

That is just multiplication of fractions.

\(\displaystyle \dfrac{(\sqrt{a})^2}{b\sqrt{a}} = \dfrac{a}{b\sqrt{a}}.\)

Square a square root and you get what?

\(\displaystyle \dfrac{a}{b\sqrt{a}} = \dfrac{a}{b\sqrt{a}} * 1.\)

Any thing equals itself times 1.

\(\displaystyle \dfrac{a}{b\sqrt{a}} * 1 = \dfrac{a}{b\sqrt{a}} * \dfrac{1\ /\ b}{1\ /\ b}.\)

Any non-zero number divided by itself equals 1.

\(\displaystyle \dfrac{a}{b\sqrt{a}}* \dfrac{1\ /\ b}{1\ /\ b} = \dfrac{a\ /\ b}{b\sqrt{a}\ / \ b}.\)

Multiplication of fractions again.

\(\displaystyle \dfrac{a\ /\ b}{b\sqrt{a}\ /\ b} = \dfrac{\dfrac{a}{b}}{\sqrt{a}}.\)

Now if you followed that every step of the way, we just proved that

\(\displaystyle \dfrac{\sqrt{a}}{b} = \dfrac{\dfrac{a}{b}}{\sqrt{a}}.\)

Nothing mysterious, just a lot of basic steps.

Very neat. But I wasn't having a problem with Dr P's expression. I could see how he got it. I guess what intrigues about what I mention in my post is that the action called for a square root to be divided by a number and instead I divided the square of the square root. So one thing was called for and I did another. Then, taking that uncalled for result and putting it over the square root of the number out of which it came....well it felt a little magicy to me.

 

[allegansveritatem]

In general, what you are describing is this: the fraction \(\displaystyle \dfrac{\sqrt{a}}{b}\) is equal to \(\displaystyle \dfrac{\frac{a}{b}}{\sqrt{a}}\). This can be proved by cross multiplication; or obtained by multiplying the numerator and denominator of the original by \(\displaystyle \dfrac{\sqrt{a}}{b}\):

\(\displaystyle \dfrac{\sqrt{a} \cdot \dfrac{\sqrt{a}}{b}}{b \cdot \dfrac{\sqrt{a}}{b}} = \dfrac{\dfrac{\sqrt{a} \cdot \sqrt{a}}{b}}{\sqrt{a}} = \dfrac{\frac{a}{b}}{\sqrt{a}}\)

Of course, this is the opposite of rationalizing the denominator ... and algebra provides many ways to rewrite any expression.

I see what you are saying. I think what catches me a little with this problem is that I did something that was not called for by the expression. It wanted the square root divided and I divided the square of the square root. I guess it just goes to show you that if you are told to do one thing and you do something else, you won't get what you're supposed to get, but the enterprise won't be a total loss--it's an ill wind that etc...

 

[Dr.Peterson]

I see what you are saying. I think what catches me a little with this problem is that I did something that was not called for by the expression. It wanted the square root divided and I divided the square of the square root. I guess it just goes to show you that if you are told to do one thing and you do something else, you won't get what you're supposed to get, but the enterprise won't be a total loss--it's an ill wind that etc...

The classic example of the thing you're talking about (as I understand it) is \(\displaystyle \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\).

I wouldn't so much say that doing the wrong thing ends up being right -- it's not right until you make it right, which requires knowing what's right in the first place. What interests me here is that the same quantity can be represented in ways that look practically opposite. This is, in fact, one reason we teach students to rationalize the denominator (that is, turn the first form into the second): so we can recognize when two expressions represent the same number.