Question

Show that 7-√5 is irrational, give that √5 is irrational

Answer 1

Answer:

7-√5 is irrational. The proof is given below

Step-by-step explanation:

Let us assume, to the contrary, that 7-√5 is rational

That is, we can find coprime a  and b (b≠ 0) such that  

7-√5 = a/b

Therefore, 7 - a/b = √5

Rearranging this equation √5 = (7b -a)/b

since a and b are integers,so (7b -a)/b is an rational.

And so √5 is rational

But this contradicts the fact that √5 is irrational.

This  contradiction  has  arisen  because  of  our  incorrect  assumption  that  7-√5 is rational.

So, we conclude that

7-√5 is irrational.

Answer 2

Answer:

It is irrational and proof is show below.

Step-by-step explanation:

We have to prove that 7 -√5 is irrational.

As we know rational numbers are those which can be written in the the form of x/y (Where x and y are integers and y ≠ 0, and they do not have a common factor).

According to above definition let's suppose that given number is rational numbers; therefore, we can write that:

$7 - \sqrt{5} = \frac{x}{y}$

Rearranging the above equation

$7 - \frac{x}{y} = \sqrt{5}$

taking LCM on left side

$\frac{7y - x}{y} = \sqrt{5}$

Now if we evaluate  numerator (7y -x) of left side of the eqution we get an integer as a result as 7,y and x are all integers. Also, the denominator of the left side equation is also an integer (as supposed in the start). Now, if we see the right side of equation i.e. √5 , we are sure that this number is not integer at all. So, as we know for an equation to be valid both sides of it should yield the same result as a result of mathematical operations; therefore, our supposition that  $7 - \sqrt{5} = \frac{x}{y}$ is wrong and it is not a rational number.