Question

State euclid's division Lemma using so that the square of any positive integer is of the form of 5m, 5m+1,5m-1​

Answer 1

EUCLID'S DIVISION LEMMA:- LEMMA IS A PROVEN STATEMENTS THAT US USED TO PROVE ANOTHER STATEMENTS.

IT CAN BE WRITE IN THE FORM OF

a=bq+r here 0< or equals to zero<b

here a represents for dividend,

b represents for divisior and

q represents for quotient and

r represents for remainder.

So it becomes:-

Dividend=Divisior×quotient+remainder.

So let's solve question:-

According to lemma:-

a=bq+r here 0<or=tor<b

so put the values here:-

a=5q+r

(i) a= 5q+0

So a= 5q

(ii) a = 5q+1

(iii) a=5q+(-1)

So a=5q-1

now squaring both sides:-

(i) a^2=5q^2

a^2=10q

Now take common here

a^2=5(2q^2)

m= 5q(2q^2)

(ii) a^2=(5q+1)^2

a^2= a^2+2ab+b^2

a^2= 5q^2+2×5q×1+1^2

a^2= 25q+10q+1

Now take common here

a^2=5(5q+2q)+1

m=(5q+2q).

(iii) a^2=(5q-1)^2

a^2=a^2-2ab+b^2

a^2=5q^2-2×5q×-1+(-1)^2

a^2=25q+10q-1

Now take common here

a^2=5(5q+2q)-1

m=(5q+2q)

Hope it helps!

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Answer 2

Step-by-step explanation:

Question : -

→ Use Euclid's Division lemma to show that the Square of any positive integer cannot be of form 5m + 2 or 5m + 3 for some integer m.

$\huge \pink{ \mid{ \underline{ \overline{ \tt Answer: -}} \mid}}$

▶ Step-by-step explanation : -

Let ‘a’ be the any positive integer .

And, b = 5 .

→ Using Euclid's division lemma :-

==> a = bq + r ; 0 ≤ r < b .

==> 0 ≤ r < 5 .

•°• Possible values of r = 0, 1, 2, 3, 4 .

→ Taking r = 0 .

Then, a = bq + r .

==> a = 5q + 0 .

==> a = ( 5q )² .

==> a = 5( 5q² ) .

•°• a = 5m . [ Where m = 5q² ] .

→ Taking r = 1 .

==> a = 5q + 1 .

==> a = ( 5q + 1 )² .

==> a = 25q² + 10q + 1 .

==> a = 5( 5q² + 2q ) + 1 .

•°• a = 5m + 1 . [ Where m = 5q² + 2q ] .

→ Taking r = 2 .

==> a = 5q + 2 .

==> a = ( 5q + 2 )² .

==> a = 25q² + 20q + 4 .

==> a = 5( 5q² + 4q ) + 4 .

•°• a = 5m + 4 . [ Where m = 5q² + 4q ] .

→ Taking r = 3 .

==> a = 5q + 3 .

==> a = ( 5q + 3 )² .

==> a = 25q² + 30q + 9 .

==> a = 25q² + 30q + 5 + 4 .

==> a = 5( 5q² + 6q + 1 ) + 4 .

•°• a = 5m + 4 . [ Where m = 5q² + 6q + 1 ] .

→ Taking r = 4 .

==> a = 5q + 4 .

==> a = ( 5q + 4 )² .

==> a = 25q² + 40q + 16 .

==> a = 25q² + 40q + 15 + 1 .

==> a = 5( 5q² + 8q + 3 ) + 1 .

•°• a = 5m + 1 . [ Where m = 5q² + 8q + 3 ] .

→ Therefore, square of any positive integer in cannot be of the form 5m + 2 or 5m + 3 .

✔✔ Hence, it is proved ✅✅.

$\huge \orange{ \boxed{ \boxed{ \mathscr{THANKS}}}}$ ]]]