1. ## 3 of 3 prove the following contradiction.

There does not exist a largest negative rational number.

Assume, that there is a largest negative rational number; call it k.

Then 2k, which is the product of two rational numbers, is a negative rational number.

Since 2k is farther from 0 than k is, we have that 2k < k, which implies that k < k/2.

Now, k/2 = 1/ 2 k is the product of two rational numbers.

so k/2 is a negative rational number that islarger than k,

resulting in a contradiction if we sub a value in for K .

Is that ok ?

I'm off to bed so forgive if I don't reply straight away but will when I wake up tomorrow cheers,

2. ## Re: 3 of 3 prove the following contradiction.

Originally Posted by bee77
There does not exist a largest negative rational number.
Assume, that there is a largest negative rational number; call it k.
Then 2k, which is the product of two rational numbers, is a negative rational number.
Since 2k is farther from 0 than k is, we have that 2k < k, which implies that k < k/2.
Now, k/2 = 1/ 2 k is the product of two rational numbers.
so k/2 is a negative rational number that islarger than k,
resulting in a contradiction if we sub a value in for K .
Is that ok ?
To echo Prof Ivey, it is very rough correct.

You should try to prove that "between any two numbers there is a rational number."
That theorem gives this "There is no largest negative number, and no smallest positive number".

3. ## Re: 3 of 3 prove the following contradiction.

Thanks Plato I'll work on it ..

4. ## Re: 3 of 3 prove the following contradiction.

Why not just multiply by 1/2 from the start?