Use a proof by contradiction to prove that the sum of two odd integers is even.
1 answer:
Answer:
Step-by-step explanation:
to prove that the sum of two odd integers is even.
Let a and b be two odd integers.
If possible assume that
, i.e. sum is a product of 2, hence even.
Since a is odd,
for some integer k.
Subtract a from a+b to get
i.e. b is a multiple of some integer l by 2
i.e. b is even.
This contradicts our assumption that both a and b are odd
Hence proved that the sum of two odd integers is even.
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Answer:
Required solution gives series (a) divergent, (b) convergent, (c) divergent.
Step-by-step explanation:
(a) Given,
To applying limit comparison test, let and
. Then,
Because of the existance of limit and the series is divergent since
where
, given series is divergent.
(b) Given,
Again to apply limit comparison test let and
we get,
Since is convergent, by comparison test, given series is convergent.
(c) Given,
. Now applying Cauchy Root test on last two series, we will get,
- \lim_{n\to \infty}|(\frac{5}{6})^n|^{\frac{1}{n}}=\frac{5}{6}=L_1
- \lim_{n\to \infty}|(\frac{1}{3})^n|^{\frac{1}{n}}=\frac{1}{3}=L_2
Therefore,
Hence by Cauchy root test given series is divergent.