Prove that x² - 1 is divisible by a where 8 is odd positive number. Do the solution in stepwise.
kvnmurty:
x^2 -1 is divisible by 8, if x is an odd positive integer.
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let x be an integer.
x² -1 = (x - 1) (x + 1)
let x be an odd positive integer. x - 1 is an even positive integer or 0 = 2 k (say)
so x + 1 is then an even positive number = 2 k + 2
x² - 1 = 4 k (k + 1)
Now k and k + 1 are consecutive integers. Hence one of them is divisible by 2. If k is even then k+1 is an odd number. If k is an odd number, then k+1 is an even number.
Hence, the product k (k+1) = 2 * [k/2] * (k+1) or k * [ (k+1)/2 ] * 2
thus x² -1 = 8 * (k/2) * (k+1) or 8 * k * [(k+1)/2]
Hence, x² - 1 is divisible by 8.
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you can use the method of proof by induction.
x² - 1 is divisible by 8 if x = 1 .
let x² - 1 be divisible by 8 for some x. Now let us see for the next off number x+2.
(x+2)² - 1
= x² + 4x + 4 - 1
= (x² - 1) + 4 (x + 1)
as x is an odd number, x +1 is an even number. Hence 4 (x+1) is divisible by 8.
Since both terms on RHS are divisible by 8., then the (x+2)² - 1 is divisible by 8.
Hence, proved by mathematical induction method.
x² -1 = (x - 1) (x + 1)
let x be an odd positive integer. x - 1 is an even positive integer or 0 = 2 k (say)
so x + 1 is then an even positive number = 2 k + 2
x² - 1 = 4 k (k + 1)
Now k and k + 1 are consecutive integers. Hence one of them is divisible by 2. If k is even then k+1 is an odd number. If k is an odd number, then k+1 is an even number.
Hence, the product k (k+1) = 2 * [k/2] * (k+1) or k * [ (k+1)/2 ] * 2
thus x² -1 = 8 * (k/2) * (k+1) or 8 * k * [(k+1)/2]
Hence, x² - 1 is divisible by 8.
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you can use the method of proof by induction.
x² - 1 is divisible by 8 if x = 1 .
let x² - 1 be divisible by 8 for some x. Now let us see for the next off number x+2.
(x+2)² - 1
= x² + 4x + 4 - 1
= (x² - 1) + 4 (x + 1)
as x is an odd number, x +1 is an even number. Hence 4 (x+1) is divisible by 8.
Since both terms on RHS are divisible by 8., then the (x+2)² - 1 is divisible by 8.
Hence, proved by mathematical induction method.
sorry bt i need it in euclid division lemma
the essence or summary of the proof is same , when you use the lemma of Euclid. Only thing is to use the language and symbols used in the lemma.
p | a means that p divides a.
Euclid's lemma says that p | a*b, then p | a or p | b or p | a*b.
ALternately, if p |a, then p| ab. --- (5)
Also, we can say that p | a/b, then pb | a. --- (6)
Euclid's lemma says that p | a*b, then p | a or p | b or p | a*b.
ALternately, if p |a, then p| ab. --- (5)
Also, we can say that p | a/b, then pb | a. --- (6)
Let x = 2 m + 1, as any odd integer is expressed in this manner. m is an integer >= 0.
x² - 1 = (2 m + 1)² - 1 = 4 m (m + 1) --- (1)
m(m-1) = (x² - 1)/4 = an integer ---- (2)
x² - 1 = (2 m + 1)² - 1 = 4 m (m + 1) --- (1)
m(m-1) = (x² - 1)/4 = an integer ---- (2)
We know that 2 | m (m+1), as one of m and m+1 is an even integer. They are consecutive integers. that is,
either 2 | m or 2 | (m+1) , hence, as per the Lemma - (5), 2 | m * (m+1)
m(m+1) | (x²-1)/4 --- (3)
Since, 2 | m(m+1) => 2 | (x² - 1)/4
=> 8 | (x²-1) by using -- (6)
either 2 | m or 2 | (m+1) , hence, as per the Lemma - (5), 2 | m * (m+1)
m(m+1) | (x²-1)/4 --- (3)
Since, 2 | m(m+1) => 2 | (x² - 1)/4
=> 8 | (x²-1) by using -- (6)
hence, the proof is done.
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