# Prove that \[2+\sqrt{5}\] is an irrational number.

Last updated date: 20th Mar 2023

•

Total views: 306k

•

Views today: 6.85k

Answer

Verified

306k+ views

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.

Recently Updated Pages

If abc are pthqth and rth terms of a GP then left fraccb class 11 maths JEE_Main

If the pthqth and rth term of a GP are abc respectively class 11 maths JEE_Main

If abcdare any four consecutive coefficients of any class 11 maths JEE_Main

If A1A2 are the two AMs between two numbers a and b class 11 maths JEE_Main

If pthqthrth and sth terms of an AP be in GP then p class 11 maths JEE_Main

One root of the equation cos x x + frac12 0 lies in class 11 maths JEE_Main

Trending doubts

What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?