Question

Prove that √18 is irrational

Answer 1

Answer:

Let us prove √18 irrational by contradiction

Let suppose that √18 is rational. It means that we have co-prime integers 'a' and 'b' (b not equal to 0)

$such \: that \: \sqrt{18} = \frac{a}{b}$

$= \sqrt[b]{18} = a$

$= 18 {b}^{2} = {a}^{2} \: (squarring \: both \: sides)$

It means that 18 is factor of a²

Hence, 18 is also factor of a by theorem–(2)

If 18 is factor of a , it means that we can write, a=18c for some integer 'c'.

Substituting value of 'a' in (1)

$= 18 {b}^{2} =324 {c}^{2}$

$b = 18 {c}^{2}$

It means that 18 is factor of b²

Hence ,18 is also factor of 'b' by theorem.......(3)

From (2) and (3) ,we can say that 18 is factor of both a and b

but 'a' and 'b' are co-prime

Therefore, our assumption was wrong. So,√18 cannot be rational number. Hence it is irrational.

⣰⣿⣿⣿⣿⣿⣿⣿⣿⣿⣦⡀ ⠀⠀⠀⢀⣾⣿⣿⣿⠿⠿⠟⠻⠿⢿⣿⣿⣿⡆ ⠀⠀⠀⢰⣿⣿⡿⠂⠀⠀⠀⠀⠀⠀⠈⠉⢻⡇ ⠀⠀⠀⠈⠿⣿⣇⣠⠤⠤⠤⢤⣀⣤⠤⠤⣺⡏ ⠀⠀⠀⠀⠐⢉⣯⠹⣀⣀⣢⡸⠉⢏⡄⣀⣯⠁ ⠀⠀⠀⠀⠡⠀⢹⣆⠀⠀⠀⣀⡀⡰⠀⢠⠖⠂ ⠀⠀⠀⠀⠀⠈⠙⣿⣿⠀⠠⠚⢋⡁⠀⡜ ⠀⠀⠀⠀⠀⠀⢸⠈⠙⠦⣤⣀⣤⣤⡼⠁ ⠀⠀⠀⠀⠀⢀⡌⠀⠀⠀⠀⠉⢏⡉ ⠀⠀⠀⣀⣴⣿⣷⣶⣤⣤⣤⣴⣾⣷⣶⣦⡀ ⢀⣴⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣄ ⠚⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠂

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