Question

Proof by Contradiction. The sum of an irrational number and a rational number is irrational.

So how do we symbolically express an irrational number?

Answer 1

id say 1/0 is irrational :)

Answer 2

What's the definition of a rational number?

Answer 3

p/q such that p and q are ints and q not 0

Answer 4

Right

Answer 5

So let x be an irrational number and y a rational number.

Let us suppose that in fact x + y is rational, then ....

Answer 6

x + y = ....

Answer 7

good, ill try to work on that angle :)

Answer 8

Let m be any irrational number, and let n be any rational number such that; m+n = p/q, q not 0.

m + \(\cfrac{p_1}{q_1}\) = \(\cfrac{p}{q}\)

m = \(\cfrac{p}{q}\) - \(\cfrac{p_1}{q_1}\)

m = \(\cfrac{p q_1 - qp_1}{q\ q_1}\) = False

maybe?

Answer 9

Yes, because by hypothesis m is irrational, yet now you have shown is rational. That is a cotradiction.

Hence it cannot be the case that if m is irrational and n is rational then m + n is rational

Therefore: if m is irrational and n is rational then m + n is irrational.

Answer 10

thanx :)

Answer 11

make that contradiction.

Answer 12

Proof by contradiction is a very important tool in mathematics.

In general, suppose you want to show that

A => B

Logically this is exactly equivalent to

not ( A and not B)

Hence if we can show that

A and not B

is false, then it must indeed be the case that not (A and not B) is true. I.e., that A => B.

Answer 13

That's exactly what we've done here with

A = m is irrational, n is rational
B = m + n is irrational

You started with the statement A, and the statement not B, and showed that

A and not B

can't be true.