23. Show that the cube of any positive integer is of the form 4m, 4m 1 or 4m 3 for some integer m.
Answers
Answer:
To prove: cube of any positive integer is of the form 4m, 4m 1 or 4m 3 for some integer m.
Solution:
Let 'a' be any positive integer.Thus,by using euclid's division lemma
⇒ a = 4q + r , where b = 4 and 0≤ r > 4
⇒ r = 0,1,2, and 3.
First, let r = 0
⇒ a = 4q +0 ⇒ a = 4q
cube on both sides,
⇒ a³ = (4q})³
or, a³ = 64q³
or, a³ = 4(16q³) ⇒ a³ = 4m (m = 16q³)
second, let r = 1
⇒ a = 4q + 1
cube on both sides,
⇒ a³ = (4q+1)³
⇒ a³ = 64q³ +1 +12q(4q+1) [∵ (x+y)³= x³+y³+3xy(x+y)]
= 64q³ +1 + 48q² +12q
⇒ a³ = 64q³ +48q+ 12q+1
or = 4(16q³+12q²+3q)+1
⇒ a³ = 4m +1 , ( m= 16q³+12q²+3q )
Third, let r = 3
⇒ a = 4q+3
cube on both sides ,
⇒ a³ = (4q+3)³
⇒ a³ = 64q³ +9 + 36q(4q+3) [∵ (x+y)³= x³+y³+3xy(x+y)]
⇒ a³ = 64q³ +9 +144q² +108q
⇒ a³ = 64q³ +144q² +108q + 27
= 64q³ +144q² +108q +24+3
= 4(16q³ +36q² +27q +6)+3
⇒ a³ = 4m +3 ,(m = 16q³ +36q² +27q +6 )
Similarly, you can do for rest of the value of 'r'
The cube of any positive integer is of the form 4m, 4m+1 or 4m +3 for some integer m.
Step-by-step explanation:
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.
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Identity used ;
∵ ( a + b )³ = a³ + b³ + 3ab ( a + b ) .
Hence, it is solved.