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 \(9\,m , 9\, m + 1\) or \(9\,m + 8.\)

 Video Solution
Real Numbers
Ex 1.1 | Question 5

Text Solution

To Prove:

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

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 \( a=b q+r, 0 \leq r \lt b\)

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 “\(m\)”) will also be a positive integer. Now, observe carefully that the cube of all the positive integers is either of the form \(9\,m\) or \(9\,m+ 1\) or \(9\,m + 1\) for some integer \(m.\)

Steps:

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

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 \(a=3q\)

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

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

Case – II. When \(a=3q+1\)

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

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

Case – III. When \(a=3q+2\)

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

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

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

  
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