Question

prove that √2 is a irrational number.​

Answer 1

Answer:

assume that √2 is irrational

√2=a/b

squaring on both sides

2=a^2/b^2

2b^2=a^2

2 divides a^2 and a

put a=2c

2 b^2=(2c) ^2

2b^2=4c^2

b^2=2c^2

therefore 2 is the common factor of a and b

but we assume that 1 is the common factor of a and b

by contradiction

√2 is irrational.

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Answer 2

Step-by-step explanation:

Let's assume √2 as rational.

then √2 = a/b where 'a' and 'b' are coprime and b≠0

so now,

b√2 = a -(i)

now squaring both sides in (i)

so, 2b² = a² - (ii)

this shows that b² divides a², therefore 'b' divides 'a'.

so this means that 'a' and 'b' have more factors other than the common factor 1.

this contradicts the fact that √2 is rational.

this contradiction has arisen due to our wrong assumption that √2 is rational.

so, √2 is irrational.