Additional Proofs: Prove each statement below using Proof by Contradiction. 1. The sum of any rational number and any irrational number is irrational. S 2. For all integers m, if m is even, then 3m+7 is odd I integer, then, 2a +3b is even 3. If a is any odd integer, and b is any even 4. Let A and B be sets from a universe U. If Bc A, then AC B

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Аноним Аноним · Answer 1 · 2019-07-05T11:22:33+02:00

Answer:

See below.

Step-by-step explanation:

1. Suppose that the sum is rational then we can write:

a/b + i = c/d where i is irrational and by definition a/b and c/d are rational.

Rearranging:

i = c/d - a/b

Now the sum on the right is rational so 'irrational' = 'rational' which is a contradiction.

So the original supposition is false and the sum must be irrational.

2. Proof of For all integers m if m is even then 3m + 7 is odd:

If m is even then 3m is even.

Suppose 3m + 7 is even, then:

3m + 7 = 2p where p is an integer.

3m - 2p = -7

But 3m and 2p are both even so their result is even and -7 is odd.

Therefore the original supposition is false because it leads to a contradiction, so 3m + 7 is odd.