Question

Show that any positive integer in the form of 4q ,4q+2,4q+4 where q is some integer.
Answer it fast guys..
50 points
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Answer 1

Correct Question :

Show that any positive even integer in the form of 4q ,4q+2,4q+4 where q is some integer.

Answer:

Let a be any positive integer. Then b = 4.

Now, by applying Euclid's Division lemma,

a=bq+r, where a is the dividend, b is the divisor, q is the quotient and r is the reminder.

So, when any number is divided by 4, there is some quotient q with a remainder (r) =0,1,2,3,4 etc.

Case I,

When r =0,

a =4q+0=4q=2(2q) (even)

Case II,

when r=1

a =4q+1=2(2q)+1 (odd)

Case III,

When r =2

a =4q+2 =2(2q)+2(even)

Case IV,

When r=3

a=4q+3 =2(2q)+3(odd)

Case V,

When r=4

a=4q+4=2(2q)+4 (even)

Thus, from Case II and IV are cancelled as value of a comes to be an odd integer.

Hence, any even positive integer is in the form of 4q, 4q+1,4q+3,4q+4..... So on, where q is any integer.

Answer 2

Step-by-step explanation:

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$\bf \huge \: Question \: \:$

Show that any positive integer in the form of 4q ,4q+2,4q+4 where q is some integer.

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$\bf \huge \: Find\: it \:$

• Show that any positive integer
• where q is some integer

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Supposed

Let a be the positive integer.

And, b = 4 .

Euclid's division lemma,

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a = 4q + r ,

for some integer q and 0 ≤ r < 4 .

°•° possible values of r is 0, 1, 2, 3 .

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putting r = 0 .

Then a = 4q .

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putting r = 1 .

Then a = 4q + 1 .

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putting r = 2

Then a = 4q + 2 .

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putting r = 3 .

then a = 4q + 3 .

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We know a is positive integer,

so a is can't 4q + 1 , or 4q + 3

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∴ HENCE

the form of 4q ,4q+2,4q+4 where q is some integer.