State true or false -3 > -13

UkusakazaL
2021-03-09
Answered

State true or false -3 > -13

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faldduE

Answered 2021-03-10
Author has **109** answers

We know that negative number which is closer to zero is bigger.
If magnitude of a negative number is less than magnitude of other negative number then negative number having less magnitude is greater. (Here, magnitude means distance of number from zero).
Here, magnitude of -3 is 3 and magnitude of -13 is 13.
We know, 3<13 .
Therefore,
-3>-13.
Hence, given statement is true.

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Can we say that $\sqrt{2}=2/(2/(2/(\dots )))$?

We have

$\varphi =1+\frac{{\textstyle 1}}{{\textstyle \varphi}}=1+\frac{{\textstyle 1}}{{\textstyle 1+\frac{{\textstyle 1}}{{\textstyle \varphi}}}}=1+\frac{{\textstyle 1}}{{\textstyle 1+\frac{{\textstyle 1}}{{\textstyle 1+\frac{1}{\varphi}}}}}=\cdots $

(with $\varphi $ being the Golden Ratio)

Which gives us the confirmed infinite fraction

$\varphi =1+\frac{{\textstyle 1}}{{\textstyle 1+\frac{{\textstyle 1}}{{\textstyle 1+\frac{{\textstyle 1}}{{\textstyle \ddots}}}}}}$

We also have

$\sqrt{2}=\frac{{\textstyle 2}}{{\textstyle \sqrt{2}}}=\frac{{\textstyle 2}}{{\textstyle \frac{{\textstyle 2}}{{\textstyle \sqrt{2}}}}}=\frac{{\textstyle 2}}{{\textstyle \frac{{\textstyle 2}}{{\textstyle \frac{{\textstyle 2}}{{\textstyle \sqrt{2}}}}}}}=\cdots $

So by analogy we can deduce that

$\sqrt{2}=\frac{{\textstyle 2}}{{\textstyle \frac{{\textstyle 2}}{{\textstyle \frac{{\textstyle 2}}{{\textstyle \frac{{\textstyle 2}}{{\textstyle \vdots}}}}}}}}$

The sequence $({a}_{n}{)}_{n\in {\mathbb{Z}}^{+}}$ such that ${a}_{1}=\sqrt{2},{a}_{n+1}=\frac{2}{{a}_{n}}$ gives $\underset{n\to +\mathrm{\infty}}{lim}{a}_{n}=\sqrt{2}$, so indeed the representation should be correct.

Everywhere on the Internet that I see a continued fraction of $\sqrt{2}$, it is

$\sqrt{2}=1+(\sqrt{2}-1)=1+\frac{{\textstyle 1}}{{\textstyle 1+\sqrt{2}}}=1+\frac{{\textstyle 1}}{{\textstyle 2+\frac{{\textstyle 1}}{{\textstyle 1+\sqrt{2}}}}}=1+\frac{{\textstyle 1}}{{\textstyle 2+\frac{{\textstyle 1}}{{\textstyle 2+\frac{{\textstyle 1}}{{\textstyle 1+\sqrt{2}}}}}}}=\cdots $

Why haven't I seen the representation $\sqrt{2}=2/(2/(2/(2/\dots )))$ mentioned, and see the above used instead? Is something wrong about my representation?

I would say it isn't mentioned because it is not useful: you cannot approximate $\sqrt{2}$ using the representation, unlike you can using the above one (which you can do by removing $\frac{1}{1+\sqrt{2}}$ n any member).

We have

$\varphi =1+\frac{{\textstyle 1}}{{\textstyle \varphi}}=1+\frac{{\textstyle 1}}{{\textstyle 1+\frac{{\textstyle 1}}{{\textstyle \varphi}}}}=1+\frac{{\textstyle 1}}{{\textstyle 1+\frac{{\textstyle 1}}{{\textstyle 1+\frac{1}{\varphi}}}}}=\cdots $

(with $\varphi $ being the Golden Ratio)

Which gives us the confirmed infinite fraction

$\varphi =1+\frac{{\textstyle 1}}{{\textstyle 1+\frac{{\textstyle 1}}{{\textstyle 1+\frac{{\textstyle 1}}{{\textstyle \ddots}}}}}}$

We also have

$\sqrt{2}=\frac{{\textstyle 2}}{{\textstyle \sqrt{2}}}=\frac{{\textstyle 2}}{{\textstyle \frac{{\textstyle 2}}{{\textstyle \sqrt{2}}}}}=\frac{{\textstyle 2}}{{\textstyle \frac{{\textstyle 2}}{{\textstyle \frac{{\textstyle 2}}{{\textstyle \sqrt{2}}}}}}}=\cdots $

So by analogy we can deduce that

$\sqrt{2}=\frac{{\textstyle 2}}{{\textstyle \frac{{\textstyle 2}}{{\textstyle \frac{{\textstyle 2}}{{\textstyle \frac{{\textstyle 2}}{{\textstyle \vdots}}}}}}}}$

The sequence $({a}_{n}{)}_{n\in {\mathbb{Z}}^{+}}$ such that ${a}_{1}=\sqrt{2},{a}_{n+1}=\frac{2}{{a}_{n}}$ gives $\underset{n\to +\mathrm{\infty}}{lim}{a}_{n}=\sqrt{2}$, so indeed the representation should be correct.

Everywhere on the Internet that I see a continued fraction of $\sqrt{2}$, it is

$\sqrt{2}=1+(\sqrt{2}-1)=1+\frac{{\textstyle 1}}{{\textstyle 1+\sqrt{2}}}=1+\frac{{\textstyle 1}}{{\textstyle 2+\frac{{\textstyle 1}}{{\textstyle 1+\sqrt{2}}}}}=1+\frac{{\textstyle 1}}{{\textstyle 2+\frac{{\textstyle 1}}{{\textstyle 2+\frac{{\textstyle 1}}{{\textstyle 1+\sqrt{2}}}}}}}=\cdots $

Why haven't I seen the representation $\sqrt{2}=2/(2/(2/(2/\dots )))$ mentioned, and see the above used instead? Is something wrong about my representation?

I would say it isn't mentioned because it is not useful: you cannot approximate $\sqrt{2}$ using the representation, unlike you can using the above one (which you can do by removing $\frac{1}{1+\sqrt{2}}$ n any member).