That was my thought, as otherwise you have to show (very trivially) that 2^m is always even and 5^n is always odd for positive integers m,n. And, for that matter, have to show an integer cannot be even and odd!
@@pablojp3498 if 5^b = 2^a then the number could be simultaneously factorized into fives and twos. But 5^b means 5*5*5... b times, the resulting number only has 5 as the prime factor because that's what made the number. 2 for example (or any other prime) is not its factor since that would mean 5^b is also equal to 5*5*5*2... etc but how can 5 turn into 2 when primes do not share a factor.
Thanks so much! These vids are helping me get through spesh units 1 and 2 and convincing me that I can continue with it in year 12. Very well explained and I understand it all
I learned this (high school symbolic logic) as the laws _modus tollens_ and _modus ponens._ That is, given any conditional known to be true, you can prove the conclusion is true by showing that the antecedent is true. Conversely, you can prove the antecedent is false by showing the conclusion to be false. (That iss more accurately the contrapositive rather than the converse.)
@@menachemsalomon Does that mean you’re old? Because I’m 15 and I prefer talking to older people than my own generation. I mean you can tell. Look at what they talk about. Feminism. TikTok. Anime. Movies. None of those interest me
@@unclegardener No, I only got 20 years on you, so not quite old. And I get what you say about talking to your peers, though tbh I do like talking about movies. Plot, theme, and underlying ideas rather than actors, actresses, and ships, of course. ;-) By the way, at 15 I had already learned some programming concepts (and languages - ever heard of BASIC?), so much of the logic, and algebra, came much easier to me.
@@menachemsalomon Yeah the part that my peers love is the actors. They always google and send pictures of the actors and actresses in the class group chat lol And no, I haven’t heard of BASIC because (luckily I took it instead of design and technology) in ICT we only learned HTML and Python I’ll try and learn some programming languages. Thx for the advice. 😊
On the board it says: " If P => Q is true then P => not Q if false." This is incorrect.. If P is a false statement then both are true. For example: "If 2 > 5 then 3=6" is a true statement.... but "if 2>5 then 3 is not equal to 6" is also true.
That's because rational numbers are numbers that can be written as a ratio between two integers. It wouldn't make sense in this case to represent rational numbers as a ratio between two other rational numbers.
I don't think this will fully answer your question cause i do wonder about it myself sometimes but that's how rational numbers are defined: a/b and a, b have to be integers I shouldn't assume things especially when it comes to mathematics but i think it's because how we learned about numbers, we start with naturals and then we learn that naturals less than zero gives us the integers and eventually we see that some divisions with two integers don't result in a integer but because it came from a ratio mathematicians decided to call those types of numbers "rational".
@@miguelfontenele221 Well, we could say that rational numbers are the ratio of any two rational numbers, but that's a circular definition. As you say, we have to define rational numbers in terms of something we've already got: the integers.
@@qwertyTRiG i see, your comment opened my eyes to something relatively obvious. You can't define the rational numbers WITH rational numbers, so you use something you already have defined which is the integers. Interesting.
You use the Fundamental Theorem of Arithmetic. Also called the unique prime factorization theorem. It states that any integer can be expressed as the product of prime numbers and that the product is unique. For example 12 is 2 x 2 x 3. 3^b is b threes multiplied together and 7^a is a sevens multiplied together. The FTA tells you they can’t be the same number.
@@aashsyed1277 The LHS will always have a factor of 3. The RHS will always not have a factor of 3. By the fundamental theorem of arithmetic, they are not the same number. QED.
If a is positive and b is negative (or vice versa), then a/b is negative, but log2(5) is positive and there is an immediate contradiction. If they are both be negative then the negatives would cancel back to positives as (-a)/(-b) = a/b, and the case is the same as in the video
@@ow7398 but we are just discussing thay whether it is rational or not. We are not interested in it's value. It can be negative or zero or positive. So, we can't go to the actual value of log2(5). We can say that it is irrational by directly seeing it's value . Then, No need to prove anything. M I right?
@@akshaychopra6679 You can never see that a number is irrational without proving it. By definition, an irrational number has a decimal value that goes infinitely without repeating. There is no way to determine this is true compared to another number which does terminate/repeat but further on. If you check 1,000,000,000,000,000 digits of an irrational number, that doesn't prove it's irrational, because for all you know it could terminate after 2,000,000,000,000,000 digits and you just didn't check far enough. The only way to know for definite that they truly are infinitely long without repeating is to prove so logically by proving that there is *no* possible fraction a/b which is exact for that number. If the number terminates, despite how long it is, we always can have something along these lines: 0.0000527 = 527/10000000. We can do the same thing for any terminating number regardless of how big the numbers in our fraction geg, so all terminating numbers are rational. Likewise, for repeating numbers. Let's say our repeating number is 0.527527527... It really doesn't matter how many numbers are in the repeating 'sequence', I've just chosen a string of 3 numbers at random to be consistent with above. We can set x = 0.527527527... Therefore 1000x = 527.527527527... 1000x - x = 99x = 527.0 x = 527/99 Thus you can see that a fraction must exist for any terminating or repeating numbers, and so irrational numbers usually requires a more direct logical proof to show that no such fraction is possible.
![จบจนได้ ! | Black Myth: Wukong #007 [ตอนจบ?]](http://i.ytimg.com/vi/DSqvWndF0sw/mqdefault.jpg)
I was today years old when I made the connection that rational numbers are called rational because they are the RATIO of two integers.
I will teach that to my future students! Very helpful.
Once you reach 2^a = 5^b, you can also observe that this violates the uniqueness of prime factorisation.
Yes
What does that mean?
That was my thought, as otherwise you have to show (very trivially) that 2^m is always even and 5^n is always odd for positive integers m,n. And, for that matter, have to show an integer cannot be even and odd!
@@pablojp3498 if 5^b = 2^a then the number could be simultaneously factorized into fives and twos. But 5^b means 5*5*5... b times, the resulting number only has 5 as the prime factor because that's what made the number. 2 for example (or any other prime) is not its factor since that would mean 5^b is also equal to 5*5*5*2... etc but how can 5 turn into 2 when primes do not share a factor.
Factually speaking, it ALSO contradicts the uniqueness of the Fundamental Theory of Arithmetic.
2:36 Proof also that the clock on the wall is out of power.
Not thorough, it could have just taken 12n hours where n ∈ ℕ
Love the daily uploads!
Thanks so much! These vids are helping me get through spesh units 1 and 2 and convincing me that I can continue with it in year 12. Very well explained and I understand it all
I learned this (high school symbolic logic) as the laws _modus tollens_ and _modus ponens._ That is, given any conditional known to be true, you can prove the conclusion is true by showing that the antecedent is true. Conversely, you can prove the antecedent is false by showing the conclusion to be false. (That iss more accurately the contrapositive rather than the converse.)
It’s always the lads without the profile pictures and their own real name who got the legendary comments
@@unclegardener High school was almost a generation ago, uncle. But I still feel like a lad.
@@menachemsalomon Does that mean you’re old? Because I’m 15 and I prefer talking to older people than my own generation. I mean you can tell. Look at what they talk about. Feminism. TikTok. Anime. Movies. None of those interest me
@@unclegardener No, I only got 20 years on you, so not quite old. And I get what you say about talking to your peers, though tbh I do like talking about movies. Plot, theme, and underlying ideas rather than actors, actresses, and ships, of course. ;-)
By the way, at 15 I had already learned some programming concepts (and languages - ever heard of BASIC?), so much of the logic, and algebra, came much easier to me.
@@menachemsalomon Yeah the part that my peers love is the actors. They always google and send pictures of the actors and actresses in the class group chat lol
And no, I haven’t heard of BASIC because (luckily I took it instead of design and technology) in ICT we only learned HTML and Python
I’ll try and learn some programming languages. Thx for the advice. 😊
The clock on the wall is broken.
fuck sake Australia's so lucky they get to be in person with this guy.
You are an amazing teacher, thank you for sharing your knowledge and helping many students around the world appreciate the beauty of maths.
lesson on logarithms, we can use small calculator to calculate.
i understood this way faster than i thought i would've 😭
Love this
On the board it says: " If P => Q is true then P => not Q if false." This is incorrect.. If P is a false statement then both are true. For example: "If 2 > 5 then 3=6" is a true statement.... but "if 2>5 then 3 is not equal to 6" is also true.
Watching this video made me interested in learning proofs.
but this wouldn't work for log base 3 5 which is irrational but at the same time we end up with 5^b=3^a where both are odd.
Which class is this?
One question: why did we establish a,b ∈ Z⁺ instead of a,b ∈ Q⁺ ?
That's because rational numbers are numbers that can be written as a ratio between two integers. It wouldn't make sense in this case to represent rational numbers as a ratio between two other rational numbers.
Any ratio of two rational numbers can also be expressed as a ratio of two integers, so you may as well express it in its simplest form.
I don't think this will fully answer your question cause i do wonder about it myself sometimes but that's how rational numbers are defined:
a/b and a, b have to be integers
I shouldn't assume things especially when it comes to mathematics but i think it's because how we learned about numbers, we start with naturals and then we learn that naturals less than zero gives us the integers and eventually we see that some divisions with two integers don't result in a integer but because it came from a ratio mathematicians decided to call those types of numbers "rational".
@@miguelfontenele221 Well, we could say that rational numbers are the ratio of any two rational numbers, but that's a circular definition. As you say, we have to define rational numbers in terms of something we've already got: the integers.
@@qwertyTRiG i see, your comment opened my eyes to something relatively obvious. You can't define the rational numbers WITH rational numbers, so you use something you already have defined which is the integers. Interesting.
2:44 Looks like that clock doesn't work.
What if a and b are both negative, x can still be positive
Seemingly we've got a typical math class full of Asians :)
Brilliant but ...WE WANT MOOOORE!!!😊😊😊
What happens if we want to involve two even or two odd numbers..like how can we prove for log 7 with base 3 ??
You would end up at 3^a = 7^b
As 3 and 7 are both prime, this cannot be true.
You use the Fundamental Theorem of Arithmetic. Also called the unique prime factorization theorem. It states that any integer can be expressed as the product of prime numbers and that the product is unique. For example 12 is 2 x 2 x 3. 3^b is b threes multiplied together and 7^a is a sevens multiplied together. The FTA tells you they can’t be the same number.
@@jeremypnet What about the number 8?
@@jeremypnet thanks 😍
@@log2306 8 is 2 x 2 x 2. Any power of 8 is only going to have 2 as its prime factors.
ty
a & b cannot be common factors too...
1:47 What the heck...? 😂
2^0=5^0
=> log2(5)=0/0
If anyone is wondering yes I am kidding.
this is much easier than all of math 8
1btc
Can you come to my school and be my maths teacher please?
LOG BASE 2 OF 6 IS IRRATIONAL
Same reason (2x3)^b =/ 2^a by the Fundamental Theory of Arithmetic
@@evanlewis2349 no if go about it , it is like 6^a=2^b but both sides are even. So how do we prove it?
@@aashsyed1277 The LHS will always have a factor of 3. The RHS will always not have a factor of 3. By the fundamental theorem of arithmetic, they are not the same number. QED.
@@evanlewis2349 I see!
What if we take a and b to be non positive integers?
If a is positive and b is negative (or vice versa), then a/b is negative, but log2(5) is positive and there is an immediate contradiction.
If they are both be negative then the negatives would cancel back to positives as (-a)/(-b) = a/b, and the case is the same as in the video
@@ow7398 got it. Thanks 👍🏻
@@ow7398 but we are just discussing thay whether it is rational or not. We are not interested in it's value. It can be negative or zero or positive. So, we can't go to the actual value of log2(5). We can say that it is irrational by directly seeing it's value . Then, No need to prove anything. M I right?
@@akshaychopra6679 we can SEE that it's irrational, but we still need to PROVE it is.
@@akshaychopra6679 You can never see that a number is irrational without proving it.
By definition, an irrational number has a decimal value that goes infinitely without repeating. There is no way to determine this is true compared to another number which does terminate/repeat but further on.
If you check 1,000,000,000,000,000 digits of an irrational number, that doesn't prove it's irrational, because for all you know it could terminate after 2,000,000,000,000,000 digits and you just didn't check far enough.
The only way to know for definite that they truly are infinitely long without repeating is to prove so logically by proving that there is *no* possible fraction a/b which is exact for that number.
If the number terminates, despite how long it is, we always can have something along these lines: 0.0000527 = 527/10000000. We can do the same thing for any terminating number regardless of how big the numbers in our fraction geg, so all terminating numbers are rational.
Likewise, for repeating numbers. Let's say our repeating number is 0.527527527... It really doesn't matter how many numbers are in the repeating 'sequence', I've just chosen a string of 3 numbers at random to be consistent with above.
We can set x = 0.527527527...
Therefore 1000x = 527.527527527...
1000x - x = 99x = 527.0
x = 527/99
Thus you can see that a fraction must exist for any terminating or repeating numbers, and so irrational numbers usually requires a more direct logical proof to show that no such fraction is possible.
When was this clip? No one was wearing masks lmao