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Show that the following Numbers are irrational(a) √17(b)3+2√5(c)1/√7
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Show that the following Numbers are irrational(a) √17(b)3+2√5(c)1/√7
**Proof that √17 is Irrational:**

To prove that √17 is irrational, we will use a proof by contradiction.

1. Assume that √17 is rational.
2. This means that √17 can be expressed as a fraction in the form a/b, where a and b are integers and b ≠ 0.
3. We can further assume that a/b is in its simplest form, meaning that a and b have no common factors other than 1.
4. Squaring both sides of the equation √17 = a/b, we get 17 = (a^2)/(b^2).
5. Rearranging the equation, we have a^2 = 17b^2.

Now, let's consider the possibilities for the integers a and b:

6. If a is even, then a^2 is also even.
7. If a^2 is even, then 17b^2 must also be even.
8. Since 17 is odd, b^2 must be even.
9. If b is even, then a and b have a common factor of 2, contradicting our assumption that a/b is in its simplest form.

Alternatively, if a is odd, then a^2 is also odd.
10. If a^2 is odd, then 17b^2 must also be odd.
11. Since 17b^2 is odd, b^2 must be odd.
12. If b is odd, then a and b have a common factor of 1, contradicting our assumption again.

In both cases, we have arrived at a contradiction, which means our assumption that √17 is rational must be false. Therefore, √17 is irrational.

**Proof that 3 + 2√5 is Irrational:**

We will once again use a proof by contradiction to show that 3 + 2√5 is irrational.

1. Assume that 3 + 2√5 is rational.
2. This means that it can be expressed as a fraction in the form a/b, where a and b are integers and b ≠ 0.
3. We can further assume that a/b is in its simplest form, with no common factors other than 1.
4. Squaring both sides of the equation 3 + 2√5 = a/b, we get (3 + 2√5)^2 = (a^2)/(b^2).
5. Expanding the equation, we have 9 + 12√5 + 20 = (a^2)/(b^2).
6. Rearranging the equation, we get 29 + 12√5 = (a^2)/(b^2).

Now, let's consider the possibilities for the integers a and b:

7. If a is even, then a^2 is also even.
8. If a^2 is even, then 29 + 12√5 must also be even.
9. Since 29 is odd, 12√5 must be odd.
10. This implies that √5 is irrational, which we have already proven.
11. Therefore, our assumption that 3 + 2√5 is rational leads to a contradiction.

Alternatively, if a is odd, then a^2 is also odd.
12. If a^2 is odd, then 29
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Show that the following Numbers are irrational(a) √17(b)3+2√5(c)1/√7
2. 1/root7 Rationalising the denominator 1/root7×root7/root7 =root7 /7To prove: 1/root7(=root7/7) is irrational Proof: Let us assume, to the contrary, that root7/7 is rational Then, root7/7=a/broot7=7a/bRHS of above eqn is integer, i.e. rational since LHS=RHS root7=rational But this contradicts the fact that root7 is irrational.therefore our assumption is wrong that 1/root7 or root7/7 is rationalTherefore 1/root7(=root7/7) is irrational.
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Show that the following Numbers are irrational(a) √17(b)3+2√5(c)1/√7
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