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Algebra

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9 Jun 2008 12:41:43 IST

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Integral function

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Prove that: $\left [ \frac{[nx]}{n} \right ] = [x]$ where is the least integer lesser than or equal to x

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SUNDEEP ALLAMRAJU

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Joined: 28 Feb 2008

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9 Jun 2008 13:11:58 IST

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Let [nx]=t then tnx< t+1t/nx< (t+1)/n.

But [t/n]t/n< [t/n]+1 and hence,[t/n]t/nx< (t+1)/n.

Also,If we show that (t+1)/n[t/n]+1 then there lies no integer between x and (t+1)/n and hence,[x]=[t/n].To show this,we have,

t/n< [t/n]+1t< n[t/n]+n which is an integer.So,tn[t/n]+n-1(t+1)/n[t/n]+1.

So,we conclude that [x]=[t/n]=[[nx]/n].

Hari Shankar

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9 Jun 2008 14:10:15 IST

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Another method is to set x = I + f where $I \in \mathbb {Z}$ and $0 \le f <1$

Hence [x] = I

So [nx] = [nI+nf] = nI + [nf]

Now, $\frac{[nx]}{n}$ = $I + \frac{[nf]}{n}$

Obviously [nf] < n

And hence $\frac{[nf]}{n}<1$

Hence $\left [ \frac{[nx]}{n} \right ]$ = I = [x]

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