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

is not a rational number, we'll use a proof by contradiction.

Assume that 2+22+2

is rational. That means it can be expressed as the ratio of two integers aa and bb where b≠0b=0 and aa and bb have no common factors other than 1:

2+2=ab2+2

=ba

Now, let's rearrange this equation to isolate 22

:

2=ab−22

=ba−2

2=a−2bb2

=ba−2b

Now, square both sides:

2=(a−2bb)22=(ba−2b)2

2=(a−2b)2b22=b2(a−2b)2

2b2=(a−2b)22b2=(a−2b)2

2b2=a2−4ab+4b22b2=a2−4ab+4b2

0=a2−4ab+2b20=a2−4ab+2b2

This equation represents a quadratic equation in terms of aa. Now, let's consider this equation modulo 2. This means we'll look at the remainders when dividing each term by 2.

0≡a2−4ab+2b2(mod2)0≡a2−4ab+2b2(mod2)

0≡a2(mod2)0≡a2(mod2)

Since the square of any integer is congruent to either 0 or 1 modulo 2, a2≡0(mod2)a2≡0(mod2) implies that aa itself must be even.

Let a=2ka=2k, where kk is an integer.

Now, substitute a=2ka=2k into the equation:

0=(2k)2−4(2k)b+2b20=(2k)2−4(2k)b+2b2

0=4k2−8kb+2b20=4k2−8kb+2b2

0=2(2k2−4kb+b2)0=2(2k2−4kb+b2)

Since 22 is a prime number, for 2(2k2−4kb+b2)2(2k2−4kb+b2) to be 00, the term inside the parentheses must be divisible by 22. But if 22 divides 2k2−4kb+b22k2−4kb+b2, then 22 divides each of its terms, including b2b2. This implies that bb is also even.

Now, if both aa and bb are even, then they have a common factor of 22, contradicting our initial assumption that aa and bb have no common factors other than 1.

Thus, our initial assumption that 2+22+2

is rational must be false. Therefore, 2+22+2

Sure, let's consider the rational number 1221 and the irrational number 22

.

The product of 1221 and 22

is:

12×2=2221×2

=22

Here, we have a rational number (1221) multiplied by an irrational number (22

), resulting in another rational number (2222

To prove that √3 – √2 and √3 + √5 are irrational, we can use proof by contradiction.

1. Let's assume that √3 – √2 is rational. This means it can be expressed as a fraction abba, where aa and bb are integers with no common factors other than 1, and b≠0b=0.

So, 3−2=ab3

−2

=ba.

Squaring both sides, we get: 3−26+2=a2b23−26

+2=b2a2 ⇒6=a2−12b2⇒6

=2b2a2−1

This implies 66

is rational. However, we know that 66 is irrational (since 6 is not a perfect square), which contradicts our assumption. Thus, 3−23−2

must be irrational.

2. Now, let's assume that 3+53

+5

2. is rational. This means it can be expressed as a fraction cddc, where cc and dd are integers with no common factors other than 1, and d≠0d=0.

So, 3+5=cd3

+5

=dc.

Squaring both sides, we get: 3+215+5=c2d23+215

+5=d2c2 ⇒15=c2−8d24d2⇒15

=4d2c2−8d2

This implies 1515

is rational. However, we know that 1515 is irrational (since 15 is not a perfect square), which contradicts our assumption. Thus, 3+53+5

must be irrational.

Therefore, both 3−23

−2 and 3+53+5