Ex.1.1 Q5 Real Numbers Solution - NCERT Maths Class 10

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Question

Use Euclid’s division lemma to show that the cube of any positive integer is of the form \(9\,\rm{m}, 9\,\rm{m} + 1\) or \(9\,\rm{m} + 8.\)

 Video Solution
Real Numbers
Ex 1.1 | Question 5

Text Solution

To Prove:

The cube of any positive integer is of the form \(\begin{align} {9\,\rm{m}, 9\,\rm{m} + 1} \end{align}\) or \(\begin{align} {9\,\rm{m} + 8.} \end{align}\)

Reasoning:

Suppose that there is a positive integer ‘\(a\)’ . By Euclid’s lemma, we know that for positive integers \(a\) and \(b,\) there exist unique integers \(q\) and \(r,\) such that \(\begin{align} a=b q+r, 0 \leq r \lt b\end{align}\)

If we keep the value of \(b = 3\), then \(0 ≤ r < 3\)  i.e. \(r = 0\) or \(1\) or \(2\) but it can’t be \(3\) because \(r\) is smaller than \(3.\) So, the possible values for \( a = 3q\) or \(3q + 1\) or \(3q + 2.\) Now, find the cube of all the possible values of \(a.\) If \(q\) is any positive integer then its cube (let’s call it as “\(\rm{m}\)”) will also be a positive integer. Now, observe carefully that the cube of all the positive integers is either of the form \(9\,\rm{m}\) or \(9\,\rm{m}+ 1\) or \(9\,\rm{m} + 1\) for some integer \(\rm{m}.\)

Steps:

Let \(\begin{align}{“a”} \end{align}\) be any positive integer and \(q=3.\)

Then, \(a=3q+r\) for some integer \(q\ge 0\) and \(0\le r<3\)

Therefore, \(a=3q\) or \(3q+1\) or \(3q+2\)

Case – I. When \(a=3q\)

\[{{\left( a \right)}^{3}}={{\left( 3q \right)}^{3~}}=27{{q}^{3}}=9{ }\left( 3{{q}^{3}} \right) =9\,\rm{m}\]

Where \(\rm{m}\) is an integer such that \(\text{m}=3{{q}^{3}}\)

Case – II. When \(a=3q+1\)

\[\begin{align}{{\left( a \right)}^{3}}&={ }{{\left( 3q{ }+{ }1 \right)}^{3}} \\{{\left( a \right)}^{3}}&={ }27{{q}^{3}}+{ }27{{q}^{2}}+9q{ }+1 \\ {{\left( a \right)}^{3}}&={ }9\left( 3{{q}^{3}}+{ }3{{q}^{2}}+q \right){ }+1 \\ {{\left( a \right)}^{3}}&={ }9\rm{m}+1 \\\end{align}\]

Where \(\rm{m}\) is an integer such that \(\text{m}=3{{q}^{3}}+{ }3{{q}^{2}}+{ }q\)

Case – III. When \(a=3q+2\)

\[\begin{align}\,{{\left( a \right)}^{3}}&={ }{{\left( 3q+2 \right)}^{3}} \\ {{\left( a \right)}^{3}}&={ }27{{q}^{3}}+{ }54{{q}^{2}}+36q{ }+{ }8 \\ {{\left( a \right)}^{3}}&={ }9\left( 3{{q}^{3}}+{ }6{{q}^{2}}+{ }4q \right){ }+{ }8 \\ {{\left( a \right)}^{3}}&={ }9\rm{m}+8 \\\end{align}\]

Where \(\rm{m}\) is an integer such that \(\text{m}=3{{q}^{3}}+{ }6{{q}^{2}}+{ }4q\)

Thus, we can see that the cube of any positive integer is of the form \(9\rm{m},\; 9{m}+ 1\) or \(9\rm{m} +8.\)

  
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