Question

show that any positive odd integers is of form 4q + 3 , where is some integer.​

Answer 1

$\huge\mathfrak\blue{Answer:}$

To Show:

Any positive odd integers is of the form 4q + 1 or 4q+3 . Where q is some integers

Solution:

Let a be any positive integer.

By Euclid's Division Lemma,

a = bq + r

So, the possible remainders are 0, 1, 2 and 3.

=> a can be of the form 4q, 4q+1, 4q+2 and 4q+3.

where q is the quotient .

a is odd and hence cannot be of the form 4q or 4q + 2 as they are even.

So, a will be in the form of 4q + 1 or

4q + 3.

Hence proved.

Answer 2

→ Answer:

→ hey user,

→ Give Question : show that any positive odd integers is of form 4q + 3 , where is some integer.​

→ Even Though we are using cases.

→ Formula used : Euclids Division Lemma

Step-by-step explanation:

→ solution : Let   ' a ' be any odd ( + ) 've integer and[  b = 4 ] .

→            ⇒ [ ∵ By Using Euclid's Division Lemma  ]

→            ⇒ [a = 4q + r   ], 0< r < 4

→             ⇒ r = 0 , 1 , 2 , 3

→              ⇒ Now , When r = 0

→              ⇒ a = 4q            [ since 4q ⇒even positive integer ]

→              ⇒ ∴ case 1 = contradicts our assumption

→             ⇒ case 1 is not true (  a ≠ 4q)

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→ Now ,case 2 :

→         ⇒  when r = 1

→         ⇒ a = 4q + 1  [ 4q + 1  = odd + ve integer ].

→        ⇒  case two is true .

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→ Case 3 :

→          ⇒ where r = 2, then a = 4q+ 2  [ But ( 4q + 2 ) = even +ve integer.

→           ⇒ It contradicts our assumption.

→          ⇒ ∴ case 3 is not true

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→ when x = 3 , a = 4q + 3 [ 4q + 3 is odd positive integer]

→ ∴ case 4 is possible.

→ [Note: + ve is a positive integer sign  and - ve is a negative integer sign]

→ Also , Hence proved by Euclids Division Lemma that the case 3 is possible.

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thank you.