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It was the time in schools, where we were taught about finding square roots of numbers. For a long time, i found the technique very difficult, so always used to find some shortcut to get rid of the tedious method. Could not find anything useful, but while using my calculator I used to observe a pattern, something as follows :

SqRoot( x ) * SqRoot ( x+1 ) = Math.round ( (10 * x + 4 )/10 , 1 )

so like

SqRoot( 2 ) * SqRoot ( 3 ) = 2.4

SqRoot( 5 ) * SqRoot ( 6 ) = 5.4

SqRoot( 23 ) * SqRoot ( 24 ) = 23.4

The only useful thing i found about the above method was to use it for finding sq root of a number quickly ( But with only 1 decimal precision ) , when some number with easy square root is available nearby ( for example 4,16, 100 )

SqRoot(99) * SqRoot(100) = 99.4

So SqRoot ( 99) = 9.94

I am not sure, if this method is already explained somewhere on wiki.. or something similar is already available, or like might be just a part of some bigger equation. But i would do like to know and curious, that why something like this "pattern" occurs. And can it be extended more, to make square root methedology even simpler ?


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Did you mean to put the rounding in the left-hand side, and should it actually be a truncation to one decimal digit? – Sam Jun 12 '13 at 11:47
yep.. Btw, in case of 2 digits, i noticed, it can be truncated to 2 decimal places.. and so on for higher digits. See 99 below – Vishwas G Jun 12 '13 at 13:00

This doesn't explain why the "pattern" occurs; it simply verifies that the pattern does occur. I hope it should shed some light on the matter anyway.

Let $\text{Math.truncate}(x, n)$ denote the truncation of $x$ to $n$ decimal digits. Then, the equation you propose (as I've understood it) is $$ \text{Math.truncate}(\sqrt{x} \cdot \sqrt{x + 1}, 1) = x + 0.4 $$

Assume that $x$ is a positive integer. Then, an equivalent way of writing the equation is $$ x + 0.4 \leq \sqrt{x} \cdot \sqrt{x + 1} < x + 0.5 $$

Because all numbers involved are positive, we can simplify this by squaring throughout the inequality: $$ (x + 0.4)^2 \leq x \cdot (x + 1) < (x + 0.5)^2 $$

This is of course equivalent to $$ x^2 + 0.8x + 0.16 \leq x^2 + x < x^2 + x + 0.25 $$

The inequality to the left ($x^2 + 0.8x + 0.16 \leq x^2 + x$) holds for positive integers $x$. Just subtract $x^2 + 0.8x$ from both sides and conclude that $0.16 \leq 0.2x$.

The inequality to the right ($x^2 + x < x^2 + x + 0.25$) obviously holds.

And so we have shown that $\text{Math.truncate}(\sqrt{x} \cdot \sqrt{x + 1}, 1) = x + 0.4$ is true for all positive integers $x$.

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hmm.. i actually was looking for some sort of logic behind relationship ( of 0.4 ) between square root of consecutive numbers. What's so special about the digit "4" in this relationship. But thanks for the derivation! – Vishwas G Jun 12 '13 at 13:59
@VishwasGagrani would say that the digit $4$ occurs because it is the greatest digit smaller than $5$. The product $n(n+1)$ of two consecutive numbers $n$ and $n+1$ is exactly $(n+0.5)^2 - 0.25$. For instance, $5 \cdot 6=30=30.25-0.25=5.5^2$. From this, we can conclude that the square root of the product $n(n+1)$ must be smaller than $n+0.5$. That is, $\sqrt{5 \cdot 6} < 5.5$. Not-so-strictly speaking, we can thus say that the decimal digit is less than $5$. Then the square root "just happens" to be large enough, so that the decimal digit is always $4$ (instead of $3$, $2$, or what have you). – Sam Jun 12 '13 at 14:13
hmm.. ok, i will try to experiment this logic with n(n+2).. and so on. Thnx! – Vishwas G Jun 12 '13 at 14:31
@VishwasGagrani You should try different bases too! – Sam Jun 12 '13 at 14:32

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