use a direct proof to show that the product of two odd numbers is odd.
Proofs: (all the nos. i used are odd)
3 x 3 = 9
5 x 9 = 45
7 x 3 = 21
Yes, but you didn't prove the statement for "all" odd integers, only the odd integers you selected.
uhm..he didn't ask for *all* odd integers..
This is Audryana's first post
"use a direct proof to show that the product of two odd numbers is odd."
That is a general statement that means 'any' two odd numbers.
ya..so...that's wat i was saying 'any' not all..
Ok, I see you have a little to learn about logic. The terms 'any', 'for each' and 'all' are used interchangably in logic and mathematics to form universals. You provided examples, or particulars as they're called, but that didn't suggest any kind of proof. It only verified the statement for a couple cases, that's all I was trying to point out.
ok....thx for telling me that...
prove the square root of 2 is irrational
Can you show me how f of X function 6X-9 Is unto or one to one.
To prove that the square root of 2 is irrational, we can use a direct proof:
Assume, for the sake of contradiction, that the square root of 2 is rational. That means it can be written as a fraction a/b, where a and b are integers with no common factors other than 1, and b is not equal to 0.
Then, we can square both sides of the equation (√2)^2 = (a/b)^2, which simplifies to 2 = a^2/b^2. Multiplying both sides by b^2 gives us 2b^2 = a^2.
Now, observe that the left-hand side of this equation, 2b^2, must be even since it is divisible by 2. Therefore, a^2 must be even as well. This implies that a must also be even, because if a were odd, then a^2 would be odd as well.
Since a is even, we can write it as a = 2k, where k is an integer. Substituting this back into the equation, we have 2b^2 = (2k)^2, which simplifies to 2b^2 = 4k^2.
Dividing both sides by 2, we get b^2 = 2k^2. Similar to before, b^2 must also be even, which means b must be even.
Now, we have found that both a and b are even, which contradicts our initial assumption that a/b is a fraction with no common factors other than 1. Therefore, our assumption that the square root of 2 is rational must be false.
Hence, we have proven that the square root of 2 is irrational.
To prove that the square root of 2 is irrational, we can use a proof by contradiction.
Assume that the square root of 2 is rational. This means it can be expressed as a fraction in the form of p/q, where p and q are integers with no common factors (except 1) and q is not equal to 0.
We can square both sides of the equation to get:
(√2)^2 = (p/q)^2
2 = p^2/q^2
2q^2 = p^2
From this equation, we can see that p^2 must be even because 2q^2 is even. This implies that p itself must be even, as the square of an odd number is odd. So we can write p as 2k, where k is an integer.
Substituting 2k for p in the equation, we get:
2q^2 = (2k)^2
2q^2 = 4k^2
q^2 = 2k^2
Now we have an equation that shows q^2 is even, which implies that q itself must be even.
However, if both p and q are even, that means p/q can be further simplified. This contradicts our initial assumption that p/q is in its simplest form.
Therefore, our initial assumption that the square root of 2 is rational must be false. Hence, the square root of 2 is irrational.