Tag: fundemental theorem of arithmetic

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Multiple choice maths real number fundemental theorem of arithmetic real numbers on number line fundamental theorem of arithmetic

Say true or false:
A positive integer is of the form $3q + 1,$ $q$  being a natural number, then you write its square in any form other than  $3m + 1$, i.e.,$ 3m $ or $3m + 2$  for some integer $m$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the positive integer $n$ is of the form $3q, 3q+1,$ and $ 3q+2$
If $n=3q$
Squaring both sides, we get,
    $=>{ n }^{ 2 }=9{ q }^{ 2 }$
    $=>{ n }^{ 2 }=3\left( { 3q }^{ 2 } \right) $
    $=>{ n }^{ 2 }=3m$, where $m=3{ q }^{ 2 }$
Now, if $n=3q+1$
    $=>{ n }^{ 2 }={ \left( 3q+1 \right)  }^{ 2 }$
    $=>{ n }^{ 2 }=9{ q }^{ 2 }+6q+1$
    $=>{ n }^{ 2 }={ 3q\left( 3q+2 \right)  }+1$
    $=>{ n }^{ 2 }=3m+1 ,$ where $  m=q\left( 3q+2 \right) $
Now, if $n=3q+2$
    $=>{ n }^{ 2 }={ \left( 3q+2 \right)  }^{ 2 }$
    $=>{ n }^{ 2 }=9{ q }^{ 2 }+12q+4$
    $=>{ n }^{ 2 }={ 3q\left( 3q+4 \right)  }+4$
    $=>{ n }^{ 2 }={ 3q\left( 3q+4 \right)  }+3+1$
    $=>{ n }^{ 2 }=3m+1$ where $m=\left( 3{ q }^{ 2 }+4q+1 \right) $
Hence, ${ n }^{ 2 }$ integer is of the form $3m$ and $3m+1$ not $3m+2$