Question

show that any positove odd integer is of the form 4m+1 or4m+3 where m is some intiger

Answer 1

Its because 4m is always even and 1 is odd and odd + even is always odd.
It will take nos. like 1, 5, 9, 13....

Same with 4m+3,
4m is always even and 3 is odd. Hence 4m + 3 is always odd.
It will take nos. like 3, 7, 11, 15.... and these nos. are the odd nos. not given by 4m + 1

Hence all odd nos. are of the form 4m + 1 or 4m + 3

Answer 2

✴✴ Hey friends!!✴✴

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✴✴ Here is your answer↓⬇⏬⤵

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▶⏩ Let ‘a’ be the any positive integers,
=> then, b= 4.

▶⏩By Euclid's Division lemma:-)

↪➡ a= bm+r. ,, 0≤ r< b. [ m= quotient].

↪➡ a=4m+r. ,, 0≤ r< 4.

→ Hence, possible values of r= 0,1,2,3.

=> Taking r=0.

↪➡ a= 4m+0 → 4m.

=> Taking r=1.

↪➡ a= 4m+1.

=> Taking r=2.

↪➡ a= 4m+2.

=> Taking r=3.

↪➡ a= 4m+3.

▶⏩ Hence, some odd integers are 4m+1 or 4m+3.

✴✴ Therefore, it is proved that 4m+1 and 4m+3 are positive odd integers for some integers m. ✴✴.


✴✴ Thanks!!✴✴.

☺☺☺ hope it is helpful for you ✌✌✌.