Show that any positive odd integer is of the form 6q + 1, or 6q + t , where q is some integer.
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Answers
Step-by-step explanation:
Using Euclid division algorithm, we know that a=bq+r, 0≤r≤b ----(1)
Let a be any positive integer and b=6.
Then, by Euclid’s algorithm, a=6q+r for some integer q≥0, and r=0,1,2,3,4,5 ,or 0≤r<6.
Therefore, a=6qor6q+1or6q+2or6q+3or6q+4or6q+5
6q+0:6 is divisible by 2, so it is an even number.
6q+1:6 is divisible by 2, but 1 is not divisible by 2 so it is an odd number.
6q+2:6 is divisible by 2, and 2 is divisible by 2 so it is an even number.
6q+3:6 is divisible by 2, but 3 is not divisible by 2 so it is an odd number.
6q+4:6 is divisible by 2, and 4 is divisible by 2 so it is an even number.
6q+5:6 is divisible by 2, but 5 is not divisible by 2 so it is an odd number.
And therefore, any odd integer can be expressed in the form 6q+1or6q+3or6q+5
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