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# Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

CBSE/Class 10/Mathematics/UNIT I: Number systems/Real numbers/NCERT Solutions/Exercise 1.1

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6. Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5 Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer 6q... read more

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm,

a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.

Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer

6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer

6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an integer

Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.

Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3,

or 6q + 5

Tutor

Let 'a' be any +ve integer. Also suppose there exists two +ve integers q and r as quotient and remainder respectively when 'a' is divided by 6. Then , By using Euclid's Division lemma, we get a = 6q + r .................( A ) , where 0 ≤ r < 6. Obviously, r = 0 or 1 or 2 or 3 or... read more

Let  'a' be any +ve integer.

Also suppose there exists two +ve integers q and r as quotient and remainder respectively when 'a' is divided by 6.

Then ,

By using Euclid's Division lemma, we get

a = 6q + r .................( A ) , where  0 ≤ r < 6.

Obviously,  r = 0 or 1 or 2 or 3 or 4 or 5.

Case-(i) When r = 0 , then from ( A ) we have

a = 6q + 0

=> a = 6q = 2×3q = 2m = Even number , where  m = 3q

Case-(ii) When r = 1 , then

a = 6q + 1 = 2(3q) + 1 = 2m + 1 = odd number, where m = 3q.

Case-(iii) When r = 2 then

a = 6q + 2 = 2(3q + 1) = 2m= Even , where m1 = 3q + 1.

Case-(iv) When r = 3 then

a = 6q + 3 = 2(3q + 1) + 1 = 2m1 + 1= odd number, where m= 3q + 1

Case-(v) When r= 4 then

a = 6q + 4 = 2(3q + 2) = 2m= Even , where m2 = 3q +2 .

Case -(vi) When r = 5 then

a = 6q + 5 = 2(3q + 2) + 1 = 2m+ 1 = odd number, where m= 3q + 2.

Clearly from above cases ( ii ) , ( iv ) and ( vi ) we conclude that any +ve odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5.

Tutor

Let a and b are two integers, where a>b, then according to Euclid division lemma a=bq+r where 0<=r<b, Here let b=6, then a=6q+r so 0<=r<6 When r=0, a=6q, when r=1,a=6q+1, for r=2, a=6q+2, for r=3, a=6q+3, for r=4, a=6q+4, for r=5, a=6q+5.here 6q,6q+2 and 6q+4 are divisible by 2, so even. 6q+1,6q+3... read more

Let a and b are two integers, where a>b, then according to Euclid division lemma a=bq+r where 0<=r<b, Here let b=6, then a=6q+r so 0<=r<6

When r=0, a=6q, when r=1,a=6q+1, for r=2, a=6q+2, for r=3, a=6q+3, for r=4, a=6q+4, for r=5, a=6q+5.here 6q,6q+2 and 6q+4 are divisible by 2, so even.

6q+1,6q+3 and 6q+5 are not divisible by 2. so they are odd.

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