Ex.1.3 Q3 Integers Solution - NCERT Maths Class 7

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Question

i) For any integer \(a,\) what is \(\left( {-1} \right) \times a\) equal to? 

ii) Determine the integer whose product with \((–1)\) is

(a) \(-22\)

(b) \(37\)

(c) \(0\)

Text Solution

Steps:

(i) \(\left( {{\rm{-1}}} \right){{ a}}\,{{ = }}\,{{-a}}\) where \(a\) is any integer

(ii) a) \({{? \times  }}\left( {{\rm{-1}}} \right){\rm{  =  -22}}\)

Let \(x\) be the required integer

\[\begin{align}x \times \left( {-1} \right) &= -22\qquad \dots\left( 1 \right)\\-x &= -22\\x &= 22\end{align}\]

Putting the value of \(x\) in equation (\(1\)), we get

\[\begin{align}x \times \left( {-1} \right) = -22\\22 \times \left( {-1} \right) = -22\end{align}\]

b) \(? × (–1) = 37\)

Let \(y\) be the required integer

\[\begin{align}y \times \left( {-1} \right)&= 37\dots(1)\\-y &= {\rm{ }}37\\y &= -{\rm{ }}37 \end{align}\]

Putting the value of \(y\) in equation (1), we get

\[–37 × (–1) = 37\]

c) \(? × (–1) =0\)

Let \(z\) be the required integer

\[\begin{align}z \times \left( {-1} \right) = 0 & \left( 1 \right)\\-z = 0\\z = 0\end{align}\]

Putting the value of z in equation (\(1\)), we get

\[\begin{align}0 × (–1) = 0\\ ? × (–1) =0 \\0 × (–1) =0 \end{align}\]

(Multiplying any number by \(0\), we get the product \(0\))

Hence (c) \(0\) is required integer.

  
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