Proof that $2^{0.5}$ will not touch the $1.5$ mark on the number line when we try to mark it exactly?

Question

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.

Answer 1

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...

Answer 2

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...

Answer 3

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

Answer 4

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.