Question

show that the of any odd integer is of the
form 4q+1 for some inteqerg.
show that 23 is an irrational q​

Answer 1

Correct question:

their should be 2m+1

$\huge{\orange{\boxed{ \pink{\mathfrak{answer}}}}}$

Let a be any +ve integer

Using Euclid's Division Lemma,

$a = bq + r \: \: \: \: \: \: \: \: \: (0 \leqslant r < b)$

Here, b = 2

$a = 2 q + r \: \: \: \: \: \: \: \: \: \: (0 \leqslant r < 2)$

For r = 0,

$a = 2q + 0 \\ = > 2q \\ \\ squaring \: both \: sides \\ \\ {a}^{2} = {(2q)}^{2} \\ {a}^{2} = 4 {q}^{2} = 2(2 {q}^{2} )$

For r = 1,

$a = 2q + 1 \\ \\ s.b.s \\ \\ {a}^{2} = {(2q + 1)}^{2} \\ {a}^{2} = 4 {q}^{2} + 1 + 2q \\ = 2(2 {q}^{2} + q) + 1$

Hence, only 2m + 1 is the only +ve odd integer.

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Answer 2

Correct Question :

Show that the of any odd integer is of the form 4q + 1 for some inteqer q.

AnswEr:

• Square of any positive integer is in form of 4q or 4q + 1 , where q is any integer.✔️

Step-By-Step ExPlanation:

Let positive integer a be the any positive integer.

Then, b = 4 .

By division algorithm we know here

0 ≤ r < 4 , So r = 0, 1, 2, 3.

When r = 0

a = 4m

Squaring both side , we get

a² = ( 4m )²

a² = 4 ( 4m²)

a² = 4q , where q = 4m²

When r = 1

a = 4m + 1

squaring both side , we get

a² = ( 4m + 1)²

a² = 16m² + 1 + 8m

a² = 4 ( 4m² + 2m ) + 1

a² = 4q + 1 , where q = 4m² + 2m

When r = 2

a = 4m + 2

Squaring both hand side , we get

a² = ( 4m + 2 )²

a² = 16m² + 4 + 16m

a² = 4 ( 4m² + 4m + 1 )

a² = 4q , Where q = 4m² + 4m + 1

When r = 3

a = 4m + 3

Squaring both hand side , we get

a² = ( 4m + 3)²

a² = 16m² + 9 + 24m

a² = 16m² + 24m + 8 + 1

a² = 4 ( 4m² + 6m + 2) + 1

a² = 4q + 1 , where q = 4m² + 6m + 2

• Hence ,Square of any positive integer is in form of 4q or 4q + 1 , where q is any integer.

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