logic's question: lets assume I have variable x which it gives me volume 5a then what's the volume of 3x?

Ryan$

Junior Member
Joined
Jan 25, 2019
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177
Hi guys, I have a question here:
lets assume I have variable x which it gives me volume 5a
then what's the volume of 3x? my question here is:
I know if I multiple x by 3 then I will get 3*5a=15a but my question who said that the relation between x and 5a is linear also who said that if x=5a then must 3x give me 15a .. is that math assumptions? any help
 

ksdhart2

Senior Member
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Mar 25, 2016
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1,053
If you had the relationship \(x = 5a\) then it would be true that \(3x = 15a\) because you multiplied one side of the equation by 3 (to get 3x) so you have to also multiply the other side by 3 (to get 15a). However, if you carefully read what you wrote that's not at all what's going on. You said:

I have variable x which it gives me volume 5a
This means that the volume (...of what?) is \(5a\) when (something?) is \(x\), but this doesn't imply an equality between \(x\) and \(5a\). Consider a cube where each side is 2 cm long. This cube would have a volume of 8 cm3. By your logic that a volume relationship implies equality, 2 and 8 should be equal. Is that true? Does that make any sense?

Similarly, consider a sphere with radius of 3 cm. This sphere would have a volume of \(\frac{4}{3} \pi (3)^3 = 36 \pi \approx 113.1\). By your logic, 3 and \(36 \pi\) should be equal. Is that true? Does that make any sense?
 

Ryan$

Junior Member
Joined
Jan 25, 2019
Messages
177
thank you very much for your explanation, very meaningful.
to sum up .. must I through my solution follow math's assumptions/rules? for example x=5a
how many 3x? I must according to math multiply x=5a by 3 so I get 3x=15a
so the answer for "how many 3x" =>15a , and that's because I used math arithmetic/rules..must I follow in every step math rules for getting correct logic?
 

HallsofIvy

Elite Member
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Jan 27, 2012
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4,877
IF you were given that "x= 5a" then, yes, multiplying on both sides by 3, "3x= 15a". But in your first post you didn't say "x= 5a". You said "I have variable x which it gives me volume 5a" which I would interpret as saying that you have some function that, applied to x, gives 5a. But there are infinitely many different functions, f, such that f(x)= 5a. One such is \(\displaystyle f(u)= \frac{5au}{x}\). When u= x, \(\displaystyle f(x)= \frac{5ax}{x}= 5a\). And for that function, \(\displaystyle f(3x)= \frac{5a(3x)}{x}= 15a\). But another such function is \(\displaystyle f(u)= \frac{5ax}{u}\). Again, \(\displaystyle f(x)= \frac{5ax}{x}= 5a\). But now \(\displaystyle f(3x)= \frac{5ax}{3x}= \frac{5}{3}a\).
 
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