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I know this would be kind of silly, but then this has been troubling me for the past few days. All of us know $2^{\frac{1}{2}}$ is irrational. Let us try to mark this on the number line "exactly", as in trying to take the first $100$ decimal places, and try to mark it through successive magnification on the number line. For every next decimal digit, it shifts to the right. It's like: $1, 1.4, 1.41, 1.414, \ldots$ Since we know that it is irrational, it keeps moving to the right infinitely. So, talking from the visual point of view, i.e. on the number line, how can we be sure that it will not touch the $1.5$ mark?

Thanks.

Sandeep
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  • As I can guess, no matter how far you go to right in the sequence. Value will always be less than 1.5. – Harish Mar 14 '15 at 15:02
  • I don't understand what you mean by "touch the 2.2 mark". Clearly $\sqrt 2$ is less than, say, $1.5$ - any representation which puts it above that is just wrong. – Joffan Mar 14 '15 at 15:03
  • @Sandeep I gave a video link; check it out. – Panglossian Oporopolist Mar 14 '15 at 15:03
  • ooops. i meant 1.5. sorry – Sandeep Mar 14 '15 at 15:04
  • Right; a looka t the video I;ve linked to will clear doubt regarding the number line; the informal proof/statement I;'ve stated below in my answer. – Panglossian Oporopolist Mar 14 '15 at 15:06
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    It keeps moving to the right infinitely, but not unboundedly. Do not confuse the infinite number of steps with the infinite extent of the accumulated steps, like in Zenon's paradox. –  Mar 14 '15 at 15:09
  • @YvesDaoust So, is there going to be a point where there is a bound where we can actually get closer to it than 1.42? – Sandeep Mar 14 '15 at 15:14
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    I mean, if an approximation to one decimal is $1.4$, you know for sure that $\sqrt2<1.5$. Any decimal you add cannot break the bound, how could it ? –  Mar 14 '15 at 15:21
  • OK. Got it people. Thanks a lot for your patience for tolerating this silly one. – Sandeep Mar 14 '15 at 15:24

4 Answers4

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The simple way is to compute $1.5^2=2.25 \gt 2$, so $\sqrt 2 \lt 1.5$

Ross Millikan
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We know this because for each decimal digit, besides the lower bound that you write, there is also an upper bound:

  • $1 < \sqrt{2} < 2$ because, squaring, we get $1 < 2 < 4$.
  • $1.4 < \sqrt{2} < 1.5$ because, squaring, we get $1.96 < 2 < 2.25$.
  • $1.41 < \sqrt{2} < 1.42$ because, squaring, we get $1.9881 < 2 < 2.0164$

And so on...

Lee Mosher
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This reminds me of Zeno's paradoxes.

Using base $10$, or any other ordinary integer base $b > 1$, if we try to mark the position of $\sqrt{2}$ (or $2^{\frac{1}{2}}$ if you prefer) with gradually improving rational approximations with a power of the base in the denominator, it is indeed the case that it keeps moving to the right infinitely.

But at each step, it moves a much smaller amount. If the next digit $d_n$ is $0$, then there's no move at all. But if the next digit is $9$ (or $b - 1$), then the next digit still adds less than $$10^{-n + 1} = \frac{1}{10^{n - 1}}$$ or $$b^{-n + 1} = \frac{1}{b^{n - 1}}$$ to our rational approximation.

Look at it in binary: $\sqrt{10} \approx 1.011010100000100111100110011$ (see Sloane's A004539). In binary, $\frac{3}{2}$ is $1.1$. But our first approximation in binary gives us $1.0$, not $1.1$. At the second step we get $1.01$. We could round this up to $1.1$ if we wanted, but that's not the procedure you're describing.

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It can be exactly marked on the number line (I.e. with nigh-infinite precision; infinite precision if pencil used has lead of size 0): https://www.youtube.com/watch?v=lmIZ8kNIQmA

It will infinitely go right; but only over a finite length... so it really doesn't eventually reach any point on the number line besides the point it is supposed to be on.

Regarding 2.2, it has bounds above and below it... and you can infinitely keep defining the rational numbers it comes between in...

$1.41 < \sqrt{2} < 1.42$

$1.413 < \sqrt{2} < 1.415$

And so on...

Panglossian Oporopolist
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  • It will infinitely go right; but only over a finite length... then, shouldn't it be rational? And I didn't get you from your 2nd paragraph quite clearly... – Sandeep Mar 14 '15 at 15:05
  • Um... no? $\pi$ is finite; 3.1415..... But it's irrational. – Panglossian Oporopolist Mar 14 '15 at 15:07
  • @Sandeep I mean to say that each irrational and rational number is represented by only one point on a number line; meaning that the 'proof' for why it not 2.2 or 1.5 is because it simply doesn't equal them; two unequal numbers can't be represented by the same point on the number line. – Panglossian Oporopolist Mar 14 '15 at 15:10