Show that cube of any positive integer is of the form 4 m. 4 m + 1 or 4 m+ 3. for some integer m.
Answers
Answer:
Step-by-step explanation:
Let a be an arbitary constant. Then by Euclid's division algorithm corresponding to the positive integers a and 4, there exists non negative integers q and r such that,
a = 4q+r where 0 ≤ r < 4
Hence, r = 0,1,2,3
Now, a³ = (4q + r)³
= 64q³+48q²r+12qr²+r³
Case 1
r = 0
64q³ = 4(16q³) = 4m, where 16q³ = m
Case 2
r = 1
= 64q³+48q²+12q+1
= 4(16q³+12q²+3q)+1
= 4m+1, where (16q³+12q²+3q) = m
Case 3
r=2
= 64q³+96q²+48q+8
= 4(16q³+24q²+12q+2)
= 4m, where (16q³+24q²+12q+2) = m
Case 4
r = 3
= 64q³+144q²+108q+27
= 64q³+144q²+108q+24+3
= = 4(16q³+36q²+27q+6)+3
= 4m+3, where (16q³+36q²+27q+6) = m
Hence, cube of any positive integer is of the form 4 m. 4 m + 1 or 4 m+ 3. for some integer m. [Proved]
Step-by-step explanation:
Let 'a' be any positive integer and b = 4.
Using Euclid Division Lemma,
a = bq + r [ 0 ≤ r < b ]
⇒ a = 3q + r [ 0 ≤ r < 4 ]
Now, possible value of r :
r = 0, r = 1, r = 2, r = 3
CASE 1:
If we take, r = 0
⇒ a = 4q + 0
⇒ a = 4q
On cubing both sides,
⇒ a³ = (4q)³
⇒ a³ = 4 (16q³)
⇒ a³ = 9m [16q³ = m as integer]
CASE 2:
If we take, r = 1
⇒ a = 4q + 1
On cubing both sides ;
⇒ a³ = (4q + 1)³
⇒ a³ = 64q³ + 1³ + 3 * 4q * 1 ( 4q + 1 )
⇒ a³ = 64q³ + 1 + 48q² + 12q
⇒ a³ = 4 ( 16q³ + 12q² + 3q ) + 1
⇒ a³ = 4m + 1 [ Take m as some integer ]
CASE 3:
If we take r = 2,
⇒ a = 4q + 2
On cubing both sides ;
⇒ a³ = (4q + 2)³
⇒ a³ = 64q³ + 2³ + 3 * 4q * 2 ( 4q + 2 )
⇒ a³ = 64q³ + 8 + 96q² + 48q
⇒ a³ = 4 ( 16q³ + 2 + 24q² + 12q )
⇒ a³ = 4m [Take m as some integer]
CASE 4 :
If we take, r = 3
⇒ a = 4q + 3
On cubing both the sides;
⇒ a³ = (4q + 3)³
⇒ a³ = 64q³ + 27 + 3 * 4q * 3 (4q + 3)
⇒ a³ = 64q³ + 24 + 3 + 144q² + 108q
⇒ a³ = 4 (16q³ + 36q² + 27q + 6) + 3
⇒ a³ = 4m + 3 [Take m as some integer]
Hence, the cube of any positive integer is in the form of 4m, 4m+1 or 4m+3.
__________________
Identity used ;
∵ ( a + b )³ = a³ + b³ + 3ab ( a + b ) .
Hence, it is solved.