Question

Show that 5√6 is a irratinoal number. ​

Answer 1

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Show that 5√6 is an irrational number.

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Let 5√6 is a rational number.

So,

5√6 = R .......(1)

√6 = $\large{\frac{R}{5}}$ .....(2)

As we know that,

• Sum of two rational number is an rational number.
• Subtraction of two rational number is an rational number.
• Multiplication of two rational number is an rational number.
• Division of two rational number is an rational number. .....(3)

As 5 is an rational number.

R is an rational number as per our assumption.

So Ratio of two rational number is also an rational number.

Means,

$\large{\frac{R}{5}}$ is a rational number.

Then,√6 is also a rational number ......(4)

But √6 is an irrational number.

So this is cause of wrong statement of equation 4).

If equation 4) is wrong, equation 2) is also wrong.

Similarly,

If equation 2) is wrong, equation 1) is also wrong.

Means our assumption is wrong.

If a real number is not rational number, it must be an irrational number.

Hence Proved

Answer 2

Answer:

Step-by-step explanation:

The most important thing to keep in mind is to know beforehand that

Let us assume that 5√6 is rational

This means that 5√6 is equal to some rational number a/b

This in turn implies that:

√6 = a/(5b)

This means that $\sqrt{6}$ is a number which is rational

Let us then assume that $\sqrt{6}$ is equal to p/q , where p and q are integers and co-primes

Therefore, after squaring both sides, we get that

6 = $p^{2} /q^{2}$

6$q^{2}$ = $p^{2}$

This means that $p^{2}$ is divisible by 6

or p is divisible by 6

Let p be some number 6x

This implies,

6$q^{2}$ = 36$x^{2}$

$q^{2}$ = 6$x^{2}$

This means that $q^{2}$ is divisible by 6

or q is divisible by 6

Therefore, p and q are not coprimes

Hence, our assumption which says that $\sqrt{6}$ is rational is false

$\sqrt{6}$ is irrational

Therefore,

The number 5$\sqrt{6}$ becomes irrational because product of a rational and irrational is always irrational.

Hope that your doubt is cleared. Happy Math!!