Tuesday 28 May 2024

From Surds, to Ab-Surds

I still use the word surd for irrational square roots, and I know there is a undercurrent in modern math education to remove what is considered "difficult"  language in the classroom.  I leave that to those still fighting in the classroom to decide, but as a historian, the term is too rich in content not to use it, and teach itHence the title of From Surds, to Ab-Surds

When I first saw the image above I thought, Oh, that's neat. I mean I know it doesn't normally work, but then I also like the crazy wrong cancellations that work, such as

If you restrict yourself to two digit numbers, there are three more of these.  For folks who want to search them out, I will give the four at the bottom of the post.  
And in case you wondered, there are three digit examples also, and onward.  Here are a couple to get you started

\( \frac{106}{625}\),  ... \( \frac{116}{464}\),  ... and for variety \( \frac{98}{392}\),  ... 

 I know you were wondering, and yes,  it goes on... 
\( \frac{1019}{5095}\),  ... 

  I know there are lots more, so if you expand on this list, send me a note.  

As I sat and tried to think of other similar "wrong" examples that work with surds, I realized it might make a really good first or second year algebra challenge.  There is nothing very difficult about the algebra itself, so it allows the problem to be setting up the algebraic structure of the arithmetic problem.  

My early thoughts quickly generated enough to recognize a pattern to generate as many as I would want, *** 
 \( \sqrt {3 \frac{3}{8}} = 3 \sqrt{ \frac{3}{8}} \) 

*** or one in higher values.  For example:

 \( \sqrt {49 \frac{49}{2400}} = 49 \sqrt{ \frac{49}{2400}} \) 

  And in general it will always work in this form::

    \( \sqrt {n \frac{n}{n^2-1}} = n \sqrt{ \frac{n}{n^2-1}} \) 

Are there other patterns that would produce fractional oddities like these?  (Send them to me.)

I was reminded by Subramanian R that there is an easy extension of these to higher powers and roots .  For instance, the first problem can be adjusted to cube roots by using 2^3 -1 in the denominator.. 

 \(\sqrt[3](2 \frac{2}{7}) = 2 \sqrt[3](\frac{2}{7}) \)

 \(\sqrt[4](2 \frac{2}{15}) = 2 \sqrt[4](\frac{2}{15}) \)

And in general, \(\sqrt[r](k\frac{k}{k^r-1})= k\sqrt[r](\frac{k}{k^r-1})\)

The four two-digit false cancellations are 

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