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thakursiddharth
@thakursiddharth
July 2021
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Prove that the following numbers are irrational.
(i) 7 - root5
(¡¡)4-3root2÷5
(¡¡¡)root5+ root3
Answers & Comments
snehitha2
Verified answer
(i) Let 7-√5 be a rational number.
A rational number can be written in the form of p/q.
7-√5 = p/q
√5 = 7-p/q
√5 = (7q-p)/q
p, q are integers then (7q-p)/q is a rational number.
Then √5 is also a rational number.
But this contradicts the fact that √5 is an irrational number.
So, our supposition is false.
Therefore, 7-√5 is an irrational number.
Hence proved.
(ii) Let (4-3√2)/5 be a rational number.
A rational number can be written in the form of p/q.
(4-3√2)/5 = p/q
4-3√2 = 5p/q
3√2 = 4-5p/q
3√2 = (4q-5p)/q
√2 = (4q-5p)/3q
p, q are integers then (4q-5p)/3q is a rational number.
Then √2 is also a rational number.
But this contradicts the fact that √2 is an irrational number.
Hence our supposition is false.
Therefore, (4-3√2)/5 is an irrational number.
Hence proved.
(iii) Let √3+√5 be a rational number.
A rational number can be written in the form of p/q.
√3+√5 = p/q
Squaring on both sides,
(√3+√5)² = (p/q)²
{√3²+√5²+2(√3)(√5)} = p²/q²
3+5+2√15 = p²/q²
8+2√15 = p²/q²
2√15 = p²/q²-8
2√15 = (p²-8q)/q
√15 = (p²-8q)/2q
p, q are integers then (p²-8q)/2q is a rational number.
Then √15 is also a rational number.
But this contradicts the fact that √15 is an irrational number.
So, our supposition is false.
Therefore, √3+√5 is an irrational number.
Hence proved.
Hope it helps ......
3 votes
Thanks 5
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Answers & Comments
Verified answer
(i) Let 7-√5 be a rational number.A rational number can be written in the form of p/q.
7-√5 = p/q
√5 = 7-p/q
√5 = (7q-p)/q
p, q are integers then (7q-p)/q is a rational number.
Then √5 is also a rational number.
But this contradicts the fact that √5 is an irrational number.
So, our supposition is false.
Therefore, 7-√5 is an irrational number.
Hence proved.
(ii) Let (4-3√2)/5 be a rational number.
A rational number can be written in the form of p/q.
(4-3√2)/5 = p/q
4-3√2 = 5p/q
3√2 = 4-5p/q
3√2 = (4q-5p)/q
√2 = (4q-5p)/3q
p, q are integers then (4q-5p)/3q is a rational number.
Then √2 is also a rational number.
But this contradicts the fact that √2 is an irrational number.
Hence our supposition is false.
Therefore, (4-3√2)/5 is an irrational number.
Hence proved.
(iii) Let √3+√5 be a rational number.
A rational number can be written in the form of p/q.
√3+√5 = p/q
Squaring on both sides,
(√3+√5)² = (p/q)²
{√3²+√5²+2(√3)(√5)} = p²/q²
3+5+2√15 = p²/q²
8+2√15 = p²/q²
2√15 = p²/q²-8
2√15 = (p²-8q)/q
√15 = (p²-8q)/2q
p, q are integers then (p²-8q)/2q is a rational number.
Then √15 is also a rational number.
But this contradicts the fact that √15 is an irrational number.
So, our supposition is false.
Therefore, √3+√5 is an irrational number.
Hence proved.
Hope it helps ......