Question

C) If a perfect square is of n- digits, than its square
root will have n/2 digits, if n is odd. that is true or false​

Answer 1

False. If n is odd, n/2 is not an integer. This contradicts the statement.

Proof

Let's prove this.

Let any natural number be $\sf{x}$. Then a perfect square is $\sf{x^2}$.

We have the digit of the perfect square.

Let's take log on that number. But in the log value, the natural part lacks the digit by 1. So we add 1 to the log value.

$\sf{log_{10}x^2+1=n}$

$\implies\sf{2log_{10}x=n-1}$

$\implies\sf{log_{10}x+1=\dfrac{n-1}{2} +1}$

$\implies\sf{log_{10}x+1=\dfrac{n+1}{2} }$

But, the digit must be a natural number. Therefore we exclude the decimal part by taking two cases.

1. n is odd

$\dfrac{n+1}{2}$ is the digit as it is natural.

2. n is even

$\dfrac{n+1}{2}$ has the decimal part. We exclude the decimal and get $\dfrac{n}{2}$ as the digit.

More information:

Why is log base 10 is used to find the digit? How is it used?

Let's take there is a positive number $\sf{k}$. We take log on that number.

$\sf{log_{10}k=(characteristic)+(decimal)}$

The below equation is after we applied log rule.

$\implies\sf{k=10^{(characteristic)+(decimal)}}$

$\implies\sf{k=10^{(characteristic)}\times10^{(decimal)}}$

However, $\sf{10^{(decimal)}}$ cannot exceed 10. So the digit of that number exactly matches with $\sf{10^{(characteristic)}}$.

After we calculate it, we find that the digit of that number is $\sf{(characteristic)+1}$.

That characteristic is related to the digit is why log base 10 is used.