To prove that the sum \( \frac{3}{4} + \sqrt{10} \) is irrational, we will assume for the sake of contradiction that it is rational and show that this leads to a contradiction.
Let’s denote the sum as:
\[ x = \frac{3}{4} + \sqrt{10} \]
Now, if \( x \) is rational, it can be expressed as:
\[ x = \frac{p}{q} \]
where \( p \) and \( q \) are integers and \( q \neq 0 \). Rearranging gives:
\[ \sqrt{10} = x - \frac{3}{4} \]
To eliminate the fraction, we can multiply both sides of the equation by \( 4q \):
\[ 4q\sqrt{10} = 4qx - 3q \]
Now, isolating \( \sqrt{10} \):
\[ \sqrt{10} = \frac{4qx - 3q}{4q} \]
Since \( x \) is assumed to be rational, both \( 4qx \) and \( -3q \) are rational numbers. Therefore, the right-hand side \(\frac{4qx - 3q}{4q}\) is also a rational number. This means that we are saying \( \sqrt{10} \) is rational.
However, we know that \( \sqrt{10} \) is irrational (as it cannot be expressed as a fraction of integers). This contradiction means our initial assumption that \( \frac{3}{4} + \sqrt{10} \) is rational must be false.
Hence, the number \( \frac{3}{4} + \sqrt{10} \) is irrational.
In summary, since \( \sqrt{10} \) is known to be irrational and the sum of a rational number \( \frac{3}{4} \) and an irrational number \( \sqrt{10} \) must also be irrational, we conclude that \( \frac{3}{4} + \sqrt{10} \) is indeed irrational.
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