Question

Prove that √10is a irrational number

Answer 1

Step-by-step explanation:

Suppose √10 is rational .

Then √10 can be expressed as a/b , where a,b are co-prime integers and b ≠ 0 .

So (√10)² = $\frac{a^2}{b^2}$

=> 10 = $\frac{a^2}{b^2}$

=> 10b² = a²                      (I)

So 10 divides a² ,

Since a is an integer , 10 divides a as well .

Let a = 10c for some integer c .

Then a² = 100c²

So from (i) we have 10b² = 100c²

=> b² = 10c²

So we see 10 divides b² .

Since b is an integer , 10 divides b as well .

Since 10 divides both a and b , it implies that a and b must have a common factor of 10 , this contradicts the fact that a,b are co-prime .

So we conclude that √10 is irrational .

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Answer 2

yes! √10 is a irrational number, which can be proved by contradictory method as follows........

we assume that√10 is a rational number

=>√10=p/q(where, p&q belong to the set of integers, q is not equal to 0)

=> now, let's cancel out all of p/q and represent it in its least form a/b

So, √10= a/b,here a&b are co- prime.ie, a&b have no common factor other than one.

=> 10= a²/b²

=> 10b² =a²...............(1)

this concludes that,10 divides a² exactly.

=> 10 divides a 'a' exactly.( as a is an integer)

Or, we can say that 10 is a factor of 'a'

which can be stated as ...

=> a= 10*c

=> a²=100c²..............(2)

By (1) & (2)

10b²= 100c²

=> b²=10c²

this concludes that, 10 divides b² exactly

=> 10 divides b ( as b is an integer)

Or, 10 is a factor of 'b'

This way,we conclude that 10 is a factor of'a'& 'b' both. That shows that 'a'&'b' are not co-prime....

where as we supposed initially, a and b are co-prime.

this contradiction conclude that our initial assumption is wrong.

Hense,√10 should be an irrational number.

[Hense Proved]

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