Mix fraction / power / HCF

Yuseph

Yuseph

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20200816_220333.jpg
Yo guys,

Whats that ? How did he come up with that HCF ? Its not explained in the book.
 
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Factoring.

[MATH]\dfrac{3^2 * 5^5 + 3^3 * 5^3}{3^4 * 5^4} = \dfrac{(3^2 * 5^3) * (5^2 + 3^1)}{(3^2 * 5^3) * (3^2 * 5^1)} =\\ \dfrac{5^2 + 3^1}{3^2 * 5^1} = \dfrac{28}{45} .................................... edited[/MATH]What you show strikes me as overly complex, but I have no idea what the context is.
 
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[MATH]\dfrac{5^2 + 3^1}{3^2 * 5^1} = \dfrac{28}{45}. \ \ \ \ \ \ \ [/MATH] Correction.
 
Subhotosh Khan said:
First:

They are showing you HCF of which numbers (you think!)​
​
What do you think the HCF should be?​
​

​
Click to expand...
I have no idea its the 1st time i see an hcf like that. Normally i deal with regular integers
 
I know only 4 methods to get the HCF
1.Euclide algorithm
2.Prime numbers calculation
3. Unroll all CF and see which one is the highest. Which implies dedication and having no social life.
4. My favorite one. Google hcf calculator.

Now i dont know why would the author mention hcf since this is a common fraction and not a common factor.
Anyway i noticed things fit perfectly in maths so I went for the hazardous way just in case.
And heres what I got.
3power2 x 5power5 = 28 125
3power3 x 5power3 = 3375
3power4 x 5power4 = 50625

HCF calculator (lol) = 1125

Then since things fit perfectly its obvious that there had to be 3 and 5 raised both at a certain power to meet 1125.

3power1 = 3
3power2 = 9
3power3 = 27
3power4 = 81
3power5 = 243
3power6 =729
3power7 no way

5power1 = 5
5power2 = 25
5power3 = 125
5power4 = 625
5power5 no way

From there i only had to divide 1125 by either 5, 25, 125 or 625 and see if it match a number in base 3 numbers. Turns out the result was 9 x 125. (3power2 x 5power4 just what the guy found)

Now i have no clue how he got to this result without resorting to a dirty method like i did
 
JeffM said:
Factoring.

[MATH]\dfrac{3^2 * 5^5 + 3^3 + 5^3}{3^4 * 5^4} = \dfrac{(3^2 * 5^3) * (5^2 + 3^1)}{(3^2 * 5^3) * (3^2 * 5^1)} =\\ \dfrac{5^2 + 3^1}{3^2 * 5^1} = \dfrac{28}{35}.[/MATH]What you show strikes me as overly complex, but I have no idea what the context is.
Click to expand...
Nice. But the only way i learned is putting common bases together, adding and substracting indexes. I dont know how to multiply and divide non identical bases yet. And you didnt really explain how you proceeded.
 
This is about the GCF or HCF of monomials. See, for example, here.

For example, the GCF of [MATH]3^2[/MATH] and [MATH]3^3[/MATH] is [MATH]3^2[/MATH] (the lower power); and the GCF of [MATH]5^5[/MATH] and [MATH]5^3[/MATH] is [MATH]5^3[/MATH]. So the GCF of [MATH]3^25^5[/MATH] and [MATH]3^35^3[/MATH] is [MATH]3^25^3[/MATH], the product of the GCFs.

We normally do this with variables:

The GCF of [MATH]x^2[/MATH] and [MATH]x^3[/MATH] is [MATH]x^2[/MATH] (the lower power); and the GCF of [MATH]y^5[/MATH] and [MATH]y^3[/MATH] is [MATH]y^3[/MATH]. So the GCF of [MATH]x^2y^5[/MATH] and [MATH]x^3y^3[/MATH] is [MATH]x^2y^3[/MATH], the product of the GCFs.
 
Yuseph said:
Nice. But the only way i learned is putting common bases together, adding and substracting indexes. I dont know how to multiply and divide non identical bases yet. And you didnt really explain how you proceeded.
Click to expand...
Fair enough. When I see something like that, my thought is to simplify it.

I see that every term contains a power of 3 so I can factor out the lowest power of 3, which is 3^2.

[MATH]\dfrac{3^2 * 5^5 + 3^3 * 5^3}{3^4 * 5^4} = \dfrac{3^2(5^5 + 3 * 5^3)}{3^2(3^2 * 5^4)}.[/MATH]
Quelque chose pour les enfants.

Now I see that every factor of 3^2 contains a power of 5 so I can factor out the lowest power of 5, which is 5^3.

[MATH]\dfrac{3^2(5^5 + 3 * 5^3)}{3^2(3^2 * 5^4)} = \dfrac{3^2 * 5^3(5^2 + 3)}{3^2 * 5^3 (3^2 * 5)}.[/MATH]
Encore quelque chose pour les enfants.

The HCF is 3^2 * 5^3. All you need to do is factor. Do not ask me why your French text goes at the task in that particular way. As a barbarian, when I see something complex, I try to simplify. But the French do like what is profound, which is why they make the best wine in the world. I am not sure whether that predilection helps explicate math.
 
Im related to French people only by the language. My culture is couscous.
I screenshot your explanation because its hard to get a good grasp of it right now. I guess thats what i should be expecting in the next chapter algebra
Anyway its clear that when theres no hcf found your method is the only way to calculate.
Im practically done wiv this chapter. Theres only this one last thing that i cant understand can you help ? Can you tell me from the picture how the guy went from one fraction to the other i cant figure out why power 3 of base 2 disappeared. Thats really the last sticking point. It seems related to your explanation though
 

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20200817_211616.jpg

Heres what i worked on today. The answer was so obvious.
All because i kept avoiding prime numbers until today
 
Yuseph said:
Im related to French people only by the language. My culture is couscous.
I screenshot your explanation because its hard to get a good grasp of it right now. I guess thats what i should be expecting in the next chapter algebra
Anyway its clear that when theres no hcf found your method is the only way to calculate.
Im practically done wiv this chapter. Theres only this one last thing that i cant understand can you help ? Can you tell me from the picture how the guy went from one fraction to the other i cant figure out why power 3 of base 2 disappeared. Thats really the last sticking point. It seems related to your explanation though
Click to expand...
Hint: 6 + (- 3) = 3
 
What they are doing is dividing both numerator and denominator by their greatest common divisor. But what they don't do is tell you how they got that greatest common divisor.

You can see that every term is a product of some power of 3 and some power of 5. The highest common power of 3 is 3^2. The highest common power of 5 is 5^3. Factor 3^2 * 5^3 out of both numerator and denominator. Then cancel. I think it is a far more straightforward process.
 
20200817_221724.jpg

Sorry guys I took the wrong picture like an idiot.
Thats what I wanted to show you. Id really appreciate if you could show me an intermediate operation before power 3 get removed magically
 
One approach would be to factor [MATH]2^2[/MATH] out of the denominator: [MATH]2^2\cdot3^3 - 2^3 = 2^2(3^3) - 2^2(2^1) = 2^2(3^3 - 2^1) = 2^2(3^3 - 2)[/MATH]
Then factor [MATH]2^2[/MATH] out of the numerator as well, and cancel (i.e. divide both by [MATH]2^2[/MATH]).

I'd probably do this very differently, but what they're doing is fine.
 
Yuseph said:
View attachment 21062

Sorry guys I took the wrong picture like an idiot.
Thats what I wanted to show you. Id really appreciate if you could show me an intermediate operation before power 3 get removed magically
Click to expand...
22 * 33 - 23 = 22 * (33 - 2)

22 from the denominator cancels out with 22 from the numerator - there is no magic involved..

You need to do these problems with "paper & pencil" - not just stare at the screen.
 
Thanks bro but that was a while ago. I became unstoppable at cancelling now i turn the thing up side down in every way I want.20200925_131005.jpg
 
Thanks bro. I took a break in the middle of the book. Just before trigonometry. I had to start Python. Im almost finished with the language now. Ill then finish with database and study more electronics. When im done wiv all that ill return to math, because all the AI robots I wanna work on require very advanced math. Integration. Calculus. Matrice. Derivatives etc. Its very heavy but in a way its good news because the heavist is it the easiest it is to oust all competitors.
This pandemic period kind of has some benefits. Sales dropped brutally but theres never been a better moment to upgrade skills.
Besides, all those hours spent on math have been beneficial not just for my electronics skills but for my focus too. My elo rank at chess went through the roof. Call me crazy if you will but the progression curb where the slope reached an extreme inclination corresponds exactly with the period where i was studying math 24 / 7. I reached 1900 (proof on my account) when i had been struggling at 1500 for months and months. It felt so weird playing with so much ease I emailed the support service about that asking them if their algorithm changed lately. They replied it hasnt changed one bit. So that was it. Its the math, what else.
Anyway I should be back in one month. 2 months at most.
 
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