I understand what irrational numbers are and how they were first proved by the ancient Greeks. My question arsis when thinking of how a length of a line can be the square root of two if by definition the magnitude of root 2 is unending. Yet there is no question that if you draw a square with sides of magnitude one that the diagonal of the square is precisely root 2 long? same question with numbers like 1/3
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1What do you mean by the 'magnitude' of root $2$? And why is it unending? – Servaes Oct 24 '21 at 14:21
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5Don't confuse the representation of a number with the number itself. In base $10$, $\frac 15=.2$ terminates. In base $2$, the same number is infinite, it's $.\overline {0011}$ to be precise. But that's just how we choose to write the thing. – lulu Oct 24 '21 at 14:27
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2the magnitude of $\sqrt 2$ is $\sqrt 2$, period. Nothing "unending" about it. The decimal representation of $\sqrt 2$ is infinite; so what? – David C. Ullrich Oct 24 '21 at 14:41
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1Does this answer your question? Visualizing the square root of 2 – JMoravitz Oct 24 '21 at 14:45
1 Answers
This is a bit like asking why, when nails are simple and useful, they are so difficult to drive with a screwdriver. This doesn't tell you much about nails; it just tells you that the screwdriver is not well-suited to driving nails.
When you say “go on forever” you are referring to the ‘decimal representation’. The representation of a number is not the number itself, but only one particular way of writing it down. It's a tool for thinking about the number. For $\frac1{10}$ and $\frac18$ it's a tool that is well-suited to the job. For $\frac13$ and $\sqrt 2$, it's less well-suited.
When we write one-eighth as “$0.125$” that's because $$\frac18 = \frac1{10} + \frac2{100} + \frac5{1000}.$$ When we try the same thing with $\frac13$, we can never make it quite add up to $\frac13$ exactly. Is that because $\frac13$ is somehow more complicated than $\frac18$? No, it's just because what we were doing, the tool we were using, is closely tied up with the number $10$—notice the $10, 100, 1000$ on the right-hand side. $\frac13$ happens to be less closely related to the number $10$ than $\frac 18$ is.
But why should we care about $10$? Not for any specially good reason, but only because we wear these meaty bodies with ten meaty prongs sticking out in one place. If we happened to have $12$ prongs instead of $10$ you might be asking why $\frac 15$ went on forever and $\frac13$ didn't. The answer would be the same: $3$ and $12$ happen to be related in a way that $5$ and $12$ aren't, just as $8$ and $10$ are related in a way that $3$ and $10$ aren't.
As you observe, the diagonal of a square is simple. Yes, and we do have a simple way to write it down: “$\sqrt 2$”, which expresses the important, simple property that it does have: if you draw a second square whose side is the diagonal of the first square, the second square will be exactly twice the size of the first square.
The decimal representation for $\sqrt2$ goes on forever, but that doesn't mean $\sqrt2$ isn't simple, it just means that decimal representation is an inconvenient way to write it, a bad tool for the job, because $\sqrt2$ is not closely related to the number $10$ in the particular way that decimal representation demands. The fact that $\sqrt 2$ is hard to write down in this one particular notation has more to do with the notation, and less about the number itself.
I hope this is some help.
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So if we used a different base system different numbers would be classified as irrational? – CatsOnAir Oct 24 '21 at 15:38
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1ok so from what I now understand is that a irrational number has a a infinite decimal approximation due to our base system but what makes the number irrational is the number is incommensurable, or in other words there is no shared measure between two integers which equals the number, Thanks so much for you help – CatsOnAir Oct 24 '21 at 15:48
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