Question

prove that √6 is irrational​

Answer 1

The following proof is a proof by contradiction.

Let us assume that

6

is rational number.

Then it can be represented as fraction of two integers.

Let the lowest terms representation be:

6

=

b

a

where b



=0

Note that this representation is in lowest terms and hence, a and b have no common factors

a

2

=6b

2

From above a

2

is even. If a

2

is even, then a should also be even.

⟹a=2c

4c

2

=6b

2

2c

2

=3b

2

From above 3b

2

is even. If 3b

2

is even, then b

2

should also be even and again b is even.

But a and b were in lowest form and both cannot be even. Hence, assumption was wrong and hence,

6

is an irrational number. •

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Answer 2

Answer:The following proof is a proof by contradiction.

Let us assume that  

6

​

 is rational number.

Then it can be represented as fraction of two integers.

Let the lowest terms representation be:  

6

​

=

b

a

​

 where b



​

=0

Note that this representation is in lowest terms and hence, a and b have no common factors

a

2

=6b

2

From above a

2

 is even. If a

2

 is even, then a should also be even.

⟹a=2c

4c

2

=6b

2

2c

2

=3b

2

From above 3b

2

 is even. If 3b

2

 is even, then b

2

 should also be even and again b is even.

But a and b were in lowest form and both cannot be even. Hence, assumption was wrong and hence,  

6

​

$\sqrt{7}$ is an irrational number.

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Step-by-step explanation: