Prove that square root of 12 is irrational.
**I don't know if I did this correctly
PF: By contrapositive, assume sqrt(12) is rational. Then there exist an a,b as integers such that a/b is written in the lowest terms, and sqrt(12)=a/b. Then by squaring both sides, 12=a^2/b^2. For a/b to be written in the lowest term a or b or both have to be odd. then a^2=12b^2 is even since an even number squared gives and even number. So, a=2n, where n is an integer, thus a^2=4n^2. So 12=4n^2/b^2. b^2 must also be even because 4n^2 can only be divided by and even number (b^2) to get the even number 12. Since both a and b are even, this contradicts at least one being odd. Therefore sqrt(12) is irrational. QED.
Could you tell me any mistakes I made, that would help alot!
Looks good. This is the method usually used to prove that √2 is irrational, but it works for any integer. You can find lots of proofs for that online. Look some up, and see how they use the same argument.
Thank you! Could you also factor sqrt(12) to be 2sqrt(3) and prove sqrt(3) to be irrational by assuming rational?
sure could. But the argument is the same. Might as well work with what you were given, in this case at least.
Your proof is almost correct, but there is one small mistake in the reasoning. When you write "b^2 must also be even because 4n^2 can only be divided by an even number (b^2) to get the even number 12," this statement is not entirely accurate.
Let's go through the proof again addressing the mistake:
PF: Assume, by contrapositive, that sqrt(12) is rational. This means there exist integers a and b, where b is not zero, such that sqrt(12) = a/b in the lowest terms.
Squaring both sides of the equation, we get 12 = (a^2)/(b^2). Now, we can rewrite this equation as a^2 = 12b^2.
Since a^2 is an integer, 12b^2 must also be an integer. This implies that b^2 is a divisor of 12. Therefore, b^2 can take the values -12, -6, -4, -3, -2, -1, 1, 2, 3, 4, 6, or 12.
Now, consider the possibilities for b:
- If b^2 is negative, then there is no integer solution for b, as the square of any real number (including integers) cannot be negative.
- If b^2 is a positive divisor of 12, then a^2 = 12b^2. Taking the square root of both sides, we have |a| = sqrt(12) * sqrt(b^2). Since the square root of 12 is positive, |a| is an integer multiple of sqrt(b^2). This implies that a is also an integer multiple of sqrt(b^2). However, since a and b are integers, it means that sqrt(b^2) must also be an integer (to cancel out the square root). This contradicts b^2 being a divisor of 12 if b is not 1 or -1.
Hence, there are no integers a and b that satisfy sqrt(12) = a/b, contradicting our initial assumption. Therefore, sqrt(12) is irrational. QED.
Overall, your proof logic is correct, but the mistake was made when you incorrectly stated that b^2 must be even. Instead, you need to consider the possibility that b^2 can take both positive and negative values as divisors of 12.