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Asked by Poonam Last Modified
Answer To prove that 3636 and 3333
are irrational numbers, we can use a proof by contradiction.
Let's assume that 3636
is rational. This means it can be expressed as a fraction in simplest form, where both the numerator and denominator are integers and the denominator is not zero.
So, let's assume 36=ab36
=ba, where aa and bb are integers with no common factors other than 1, and b≠0b=0.
Now, let's square both sides of the equation to eliminate the square root:
Now, multiply both sides by b2b2 to clear the fraction:
So, a2a2 must be divisible by 54. This implies aa must be divisible by 5454
.
However, 54=2×3354=2×33. Since there's a 3333 term, for a2a2 to be divisible by 3333, aa must also be divisible by 33.
Now, let's consider the original equation again:
If aa is divisible by 33, then abba is also divisible by 33, but then 3636
is not in simplest form, which contradicts our assumption. Therefore, 3636
cannot be rational.
Similarly, we can show that 3333
is also irrational by following a similar proof by contradiction. Therefore, both 3636 and 3333
are irrational numbers.
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