Question

show that a(a^2+2)/3 is an integer for all    a  >   1 or equal 1 ?

Answer 1

Q. Show that [ a( a² + 2 ) ] / 3 is an integer for a ≥ 1.

To show this , first of all we need to know something about integer.

The group of positive and negative numbers collectively is known as Integets except fractional form.


Now, we can prove it by Euclid's Division Lemma.

How can we prove it using Euclid's Division Lemma ?

If [ a( a² + 3 ) ] is completely divisible by 3 , then it will be an integer.

i.e. [ a( a² + 3 ) ] /3 will be an integer.

Here we go,

Let, b = 3 as [ a( a² + 3 ) ] is going to be divided by 3.

By Euclid's Division Lemma,

=> a = bq + r

=> a = 3q + r ( 0 ≤ r ≤ 3 )

Possible values of r = 0 , 1 and 2.

Case ( i ),

When, r = 0

=> a = 3q + r

=> a = 3q + 0

•°• a = 3q

Squaring both sides,

=> a² = ( 3q )²

•°• a² = 9q²

Now,

= [ a( a² + 2 ) ] /3

Substitute the value of a and a²,

= [ 3q( 3q² + 2 ) ] /3

= q( 3q² + 2 )

Hence, a( a² + 2 ) is completely divisible by 3.

So, it's an integers.

Case ( ii ),

When , r = 1

=> a = 3q + r

=> a = 3q + 1

Squaring both sides,

=> a² = ( 3q + 1 )²

=> a² = ( 3q )² + 1² + 2 × 3q × 1

•°• a² = 9q² + 1 + 6q

Now,

= [ a( a² + 2 ) ] / 3

Substitute the value of a and a²,

= [ ( 3q + 1 ) ( 9q² + 1 + 6q + 2 ) ] / 3

= [ ( 3q + 1 ) ( 9q² + 3 + 6q ) ] / 3

= [ 3( 3q + 1 ) ( 3q² + 1 + 2q ) ] / 3

= ( 3q + 1 ) ( 3q² + 2q + 1 )

Since, a( a² + 2 ) is completely divisible by 3.

So, [ a( a² + 2 ) ] / 3 is an integer.

Case ( iii ),

When, r = 2

=> a = 3q + r

=> a = 3q + 2

Squaring both sides,

=> a² = ( 3q + 2 )²

=> a² = ( 3q )² + 2² + 2 × 3q × 2

=> a² = 9q² + 4 + 12q

Now,

= [ a( a² + 2 ) ] / 3

Substitute the value of a and a²,

= [ ( 3q + 2 ) ( 9q² + 4 + 12q + 2 ) ] / 3

= [ ( 3q + 2 ) ( 9q² + 12q + 6 ) ] / 3

= [ 3 ( 3q + 2 ) ( 3q² + 4q + 2 ) ] / 3

= ( 3q + 2 ) ( 3q² + 4q + 2 )

Since, [ a( a² + 2 ) ] is divisible by 3.

So, [ a( a² + 2 ) ] / 3 is an integer.

In all three cases , we have proved that [ a( a² + 2 ) ] / 3 is an integer.

Hope it helps !!

Answer 2

a(a²+2)/3 is an integer for all a ≥ 1 is true.

GIVEN: a(a²+2)/3

             a ∈ Z

TO PROVE: a(a²+2)/3 is an integer for all a ≥ 1.

SOLUTION:

As we know,

We will use the Principle of Mathematical Induction to prove the statement to be true.

Step 1:

Putting a = 1 in a(a²+2)/3

= 1(1+2)/3

=1

Therefore, the statement is true for a=1.

Now,

We assume that the statement is true for a = k

Step 2:

K(k²+2)/3= z.                                              [ z is an integer ]  -------eq 1

Now,

For step 3 we take a = k+1

Step 3:put a=k+1

k+1((k+1)²+2)/3

=k+1(k²+1+2k+2)/3

= k+1(k²+2k+3)/3

Now, we replace the value of k from 2 from Eq 1

Therefore,

We get,

an (a²+2)/3 is an integer for all a ≥ 1

Hence Proved.

The statement is true by the Principle of Mathematical Induction that all a belong to Z.

#SPJ2

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