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Hint: Prove that \[\sqrt{5}\] is an irrational number by assuming that it is a rational number and then using contradiction to prove that it is an irrational number. Use the fact that the sum of a rational number and an irrational number is an irrational number to prove that \[2+\sqrt{5}\] is an irrational number.

We have to prove that \[2+\sqrt{5}\] is an irrational number. We will firstly prove that \[\sqrt{5}\] is an irrational number. We will prove this by contradiction technique.

Let’s assume that \[\sqrt{5}\] is a rational number. We know that any rational number can be written in the form \[\dfrac{a}{b}\] where \[a\] and \[b\] are co-prime numbers and \[b\ne 0\].

Thus, we have \[\sqrt{5}=\dfrac{a}{b}\]. Squaring on both sides, we get \[5=\dfrac{{{a}^{2}}}{{{b}^{2}}}\].

Rearranging the terms, we get \[{{a}^{2}}=5{{b}^{2}}......\left( 1 \right)\].

We see that \[5\] divides \[{{a}^{2}}\]. We know that if a prime number \[p\] divides \[{{a}^{2}}\], then \[p\] must divide \[a\] as well, where \[a\] is a positive integer.

Thus, as \[5\] divides \[{{a}^{2}}\], \[5\] must divide \[a\] as well.

Hence, we can write \[a=5c\] for some positive integer \[c\].

Substituting the value \[a=5c\] in equation \[\left( 1 \right)\], we have \[25{{c}^{2}}=5{{b}^{2}}\Rightarrow {{b}^{2}}=5{{c}^{2}}\].

As \[{{b}^{2}}=5{{c}^{2}}\], we observe that \[5\] divides \[{{b}^{2}}\]. So, \[5\] divides \[b\] as well using the fact that if a prime number \[p\] divides \[{{a}^{2}}\], then \[p\] must divide \[a\] as well, where \[a\] is a positive integer.

Thus, \[5\] divides both \[a\] and \[b\]. But this contradicts our assumption that \[a\] and \[b\] are coprimes.

Hence, our assumption is wrong that \[\sqrt{5}\] is a rational number.

We conclude that \[\sqrt{5}\] is an irrational number.

We can write \[2\] as \[\dfrac{2}{1}\], thus observing that it is a rational number.

We know that a sum of a rational number and an irrational number is an irrational number.

Hence, we observe that \[2+\sqrt{5}\] is an irrational number.

Note: It’s necessary to keep the definition of rational numbers in mind which states that any rational number can be written in the form \[\dfrac{a}{b}\] where \[a\] and \[b\] are co-prime numbers and \[b\ne 0\], while irrational numbers are all those real numbers which are not rational numbers. Also, it’s necessary to use the theorem which states that if a prime number \[p\] divides \[{{a}^{2}}\], then \[p\] must divide \[a\] as well, where \[a\] is a positive integer. We also observe that the square root of any prime number will be an irrational number.

We have to prove that \[2+\sqrt{5}\] is an irrational number. We will firstly prove that \[\sqrt{5}\] is an irrational number. We will prove this by contradiction technique.

Let’s assume that \[\sqrt{5}\] is a rational number. We know that any rational number can be written in the form \[\dfrac{a}{b}\] where \[a\] and \[b\] are co-prime numbers and \[b\ne 0\].

Thus, we have \[\sqrt{5}=\dfrac{a}{b}\]. Squaring on both sides, we get \[5=\dfrac{{{a}^{2}}}{{{b}^{2}}}\].

Rearranging the terms, we get \[{{a}^{2}}=5{{b}^{2}}......\left( 1 \right)\].

We see that \[5\] divides \[{{a}^{2}}\]. We know that if a prime number \[p\] divides \[{{a}^{2}}\], then \[p\] must divide \[a\] as well, where \[a\] is a positive integer.

Thus, as \[5\] divides \[{{a}^{2}}\], \[5\] must divide \[a\] as well.

Hence, we can write \[a=5c\] for some positive integer \[c\].

Substituting the value \[a=5c\] in equation \[\left( 1 \right)\], we have \[25{{c}^{2}}=5{{b}^{2}}\Rightarrow {{b}^{2}}=5{{c}^{2}}\].

As \[{{b}^{2}}=5{{c}^{2}}\], we observe that \[5\] divides \[{{b}^{2}}\]. So, \[5\] divides \[b\] as well using the fact that if a prime number \[p\] divides \[{{a}^{2}}\], then \[p\] must divide \[a\] as well, where \[a\] is a positive integer.

Thus, \[5\] divides both \[a\] and \[b\]. But this contradicts our assumption that \[a\] and \[b\] are coprimes.

Hence, our assumption is wrong that \[\sqrt{5}\] is a rational number.

We conclude that \[\sqrt{5}\] is an irrational number.

We can write \[2\] as \[\dfrac{2}{1}\], thus observing that it is a rational number.

We know that a sum of a rational number and an irrational number is an irrational number.

Hence, we observe that \[2+\sqrt{5}\] is an irrational number.

Note: It’s necessary to keep the definition of rational numbers in mind which states that any rational number can be written in the form \[\dfrac{a}{b}\] where \[a\] and \[b\] are co-prime numbers and \[b\ne 0\], while irrational numbers are all those real numbers which are not rational numbers. Also, it’s necessary to use the theorem which states that if a prime number \[p\] divides \[{{a}^{2}}\], then \[p\] must divide \[a\] as well, where \[a\] is a positive integer. We also observe that the square root of any prime number will be an irrational number.

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