Step-by-step explanation:
We will prove this by contradiction. So, assuming the statement (√7 + √11) is irrational to be true.
Let k = √7 + √11
√11 = k - √7
Squaring both sides, we get,
11 = (k - √7)2
11 = k2 + (√7)2 – 2 × √7 × k
k2 + 7 - 2√7k = 11
2√7k = 11 – 7 – k2
2√7k = 4 – k2
√7
As, k is a rational number,
is also n rational number.
But √7 is an irrational number.
So, there is a contradiction.
Therefore, statement is wrong.
Hence, √7 + √11 is an irrational number.
OR
Any odd integer is of the form (2q + 1) for some integer q.
Let a = 2m + 1 and b = 2n + 1
Therefore,
a2 + b2 = (2m + 1)2+ (2n + 1)2
a2 + b2 = 4m2 + 4m + 1 + 4n2 + 4n + 1
a2 + b2 = 4{(m2 + n2) + (m + n)} + 2
Let k = {(m2 + n2) + (m + n)}
Then,
a2 + b2 = 4k + 2
a2 + b2 = 2(2k + 1)
Therefore, a2 + b2 = even number.
As the multiple is of 2 and not 4. The number is not divisible by 4.
Hence, Proved
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Answers & Comments
Step-by-step explanation:
We will prove this by contradiction. So, assuming the statement (√7 + √11) is irrational to be true.
Let k = √7 + √11
√11 = k - √7
Squaring both sides, we get,
11 = (k - √7)2
11 = k2 + (√7)2 – 2 × √7 × k
k2 + 7 - 2√7k = 11
2√7k = 11 – 7 – k2
2√7k = 4 – k2
√7
As, k is a rational number,
is also n rational number.
But √7 is an irrational number.
So, there is a contradiction.
Therefore, statement is wrong.
Hence, √7 + √11 is an irrational number.
OR
Any odd integer is of the form (2q + 1) for some integer q.
Let a = 2m + 1 and b = 2n + 1
Therefore,
a2 + b2 = (2m + 1)2+ (2n + 1)2
a2 + b2 = 4m2 + 4m + 1 + 4n2 + 4n + 1
a2 + b2 = 4{(m2 + n2) + (m + n)} + 2
Let k = {(m2 + n2) + (m + n)}
Then,
a2 + b2 = 4k + 2
a2 + b2 = 2(2k + 1)
Therefore, a2 + b2 = even number.
As the multiple is of 2 and not 4. The number is not divisible by 4.
Hence, Proved