Can any Infinity be One

Algebra Level 2

If $X + 1 = 2$ where $X = 0. YYYY\ldots$ , find the positive square root of $Y$ .

The answer is 3.

6 solutions

Saharsh Anand
Jun 28, 2015

$X+1=2 \implies X=1$

And, $X=0.YYYYYY...$ . Since, $1=0.999999...$ . Hence, $Y=9$ .

FInally, $\sqrt{Y}=\sqrt{9}=3$

Very clever solution, upvoted

Syed Baqir - 5 years, 11 months ago

a clever solution indeed

Ashish Garg - 5 years, 11 months ago

How 0.9999..... =1 it is opproximate value u know

Balaji T.S - 5 years, 11 months ago

Nope. It's exact. 0.999... =1. Proof: Let x = 0.999... 10x =9.999... 10x - x = 9.999... - 0.999... 9x = 9

Daren Lobo - 5 years, 11 months ago

What will happen if you round off 0.99999, Surely it is approximately equal to 1

Syed Baqir - 5 years, 11 months ago

It's not approximated, it's exact. It's a different way of writing one

Alex Neiman - 5 years, 11 months ago

0.99999... is really equal to 1.. 0.99999 can be written as sum of 9/10+9/100+9/1000+...with r= 1/10,hence by using the formula of sum of n terms of geometric sequence we have 9/10 divided by 1-1/10 which is equal to 1..

parson formentera - 5 years, 10 months ago

Simple approach

Rama Devi - 5 years, 11 months ago

Ikkyu San
Jun 28, 2015

From $X+1=2\Rightarrow X=1$

Thus,

$\begin{aligned}0.YYYY\ldots=&1\\\frac{Y}{10}+\frac{Y}{10^2}+\frac{Y}{10^3}+\frac{Y}{10^4}+\ldots=&1\\\dfrac{\frac{Y}{10}}{1-{\frac1{10}}}=&1\\\frac{Y}{10}=&\frac9{10}\\Y=&9\end{aligned}$

Hence, square root of $Y$ is $\boxed{3}$ .

Here is how I did
let z=.yyyyy...
10z=y.yyyyy...
9z= y
Z=1 from the eq
Y=9
Sq root for 9 equals 3

Praful Jain - 5 years, 11 months ago

@Praful Jain , The answer to this problem is incorrect. The square roots of 9 are $+/- 3$ So, Kindly rephrase your problem asking "What is the positive Square root of Y"

Thanks!

Mehul Arora - 5 years, 11 months ago

No, you don't need to specify the word "positive", when taking the square root, we are only taking the non-negative value. Unless you rephrase your question to $X^2 = 9$ , then $X = \pm 3$ .