Question

prove that 3 - 4 root 7 is irrational
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Answer 1

Step-by-step explanation:

Assume that 3-4√7 is rational

3-4√7=p/q

-4√7=p/q-3

√7=p/q-3 upon - 4

√7=3-p/q upon 4

3-p/q upon 4 is rational.

√7 is irrational.

so, our assumption is wrong.

Hence, 3 - 4 root 7 is rational.

Answer 2

Answer:

$3 - 4 \sqrt{7}$

Let us assume to the contrary that 3-4root7 is rational.

$3 - 4\sqrt{7} = \frac{a}{b}$

Where a and b are integers, a and b are coprimes, b is not equal to 0.

$3 - 4 \sqrt{7} = \frac{a}{b} \\ - 4 \sqrt{7} = \frac{a}{b} - \frac{3}{1} \\ - 4 \sqrt{7} = \frac{a - 3b}{b} \\ \sqrt{7} = \frac{a - 3b}{ - 4b} \\ \frac{a - 3b}{ - 4b}$

Since a and b are integers, it is rational

Therefore, root7 is rational.

But we know that root7 is irrational

Our assumption is wrong that 3-4root7 is rational.

Therefore, conclude 3-4root7 is irrational

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