Question

Use Euclid's division hemma, to show that the
any positive integer is of the form
5p or5p +1 or 5p +3 0r 5p +4 for any integer ‘p’

Answer 1

Answer:

not hemma its lemma

Step-by-step explanation:

using formula

a=bq+r

where r≤0<1

Answer 2

Answer:

Step-by-step explanation:

Let 'a' be any positive integer and b = 5.

Using Euclid Division Lemma,
a = bq + r            [ 0 ≤ r < b ]
⇒ a = 5q + r       [ 0 ≤ r < 5 ]

Now, possible value of r :
r = 0, r = 1, r = 2, r = 3, r = 4

CASE I :

If we take, r = 0
⇒ a = 5q + 0
⇒ a = 5q
On squaring both sides;
⇒ a² = (5q)²
⇒ a² = 25q²
⇒ a² = 5 ( 5q² )
⇒ a² = 5m.    [ Here, m = 5q² ]

CASE II :

If we take, r = 1
⇒ a = 5q + 1
On squaring both sides;
⇒ a² = (5q + 1)²
⇒ a² = (5q)² + 2 * 5q * 1 + 1²
⇒ a² = 25q² + 10q + 1
⇒ a² = 5 ( 5q² + 2q ) + 1
⇒ a² = 5m + 1   [ Here, m = 5q² + 2q ]

CASE III :

If we take, r = 2
⇒ a = 5q + 2
On squaring both sides;
⇒ a² = (5q + 2)²
⇒ a² = (5q)² + 2 * 5q * 2 + 2²
⇒ a² = 25q² + 20q + 4
⇒ a² = 5 ( 5q² + 4q ) + 4
⇒ a² = 5m + 4  [ Here, m = 5q² + 4q ]

CASE IV :

If we take, r = 3
⇒ a = 5q + 3
On squaring both sides;
⇒ a² = (5q + 3)²
⇒ a² = (5q)² + 2 * 5q * 3 + 3²
⇒ a² = 25q² + 30q + 5 + 4
⇒ a² = 5 ( 5q² + 6q + 1) + 4
⇒ a² = 5m + 4  [ Here, m = 5q² + 6q + 1 ]

CASE V :

If we take, r = 4
⇒ a = 5q + 4
On squaring both sides;
⇒ a² = (5q + 4)²
⇒ a² = (5q)² + 2 * 5q * 4 + 4²
⇒ a² = 25q² + 40q + 15 + 1
⇒ a² = 5 ( 5q² + 8q + 3 ) + 1
⇒ a² = 5m + 1  [ Here, m = 5q² + 8q + 3 ]

Hence, the square of any integer is either of the form 5m, 5m+1 or 5m+4 for some integer m.

                                       Proved.