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Math Help - Base 5 Help

  1. #1
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    Base 5 Help

    Can someone help me understand changing between bases?

    For example, if we have

    (34)_5 + (23)_5, I think of it this way:

    We have 5 5's and then 7 1's so we have 6 5's and 2 1's which means we have:

    1 (5^2) .. 1 (5^1) .. 2 (5^0)

    So our answer then would be (112)_5?

    Alternatively, can we just say (34)_5 + (23)_5 = (57)_5? I'm assuming not, since we're using base 10 rules in adding.

    Instead, I think it would be (34)_5 + (23)_5 = (62)_5

    Some advice?
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  2. #2
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    Quote Originally Posted by DiscreteW View Post
    Can someone help me understand changing between bases?

    For example, if we have

    (34)_5 + (23)_5, I think of it this way:

    We have 5 5's and then 7 1's so we have 6 5's and 2 1's which means we have:

    1 (5^2) .. 1 (5^1) .. 2 (5^0)

    So our answer then would be (112)_5?

    Alternatively, can we just say (34)_5 + (23)_5 = (57)_5? I'm assuming not, since we're using base 10 rules in adding.

    Instead, I think it would be (34)_5 + (23)_5 = (62)_5

    Some advice?
    Do you mean \log_5 (34) + \log_5 (23)? From the usual log rules, it's equal to \log_5 [(34)(23)].

    And \log_5 [(34)(23)] = x \Rightarrow 5^x = (34)(23).
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Do you mean \log_5 (34) + \log_5 (23)? From the usual log rules, it's equal to \log_5 [(34)(23)].

    And \log_5 [(34)(23)] = x \Rightarrow 5^x = (34)(23).
    Nope I mean exactly what I said.

    (34)_5 + (23)_5\ldots
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  4. #4
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    Hello,
    Quote Originally Posted by DiscreteW View Post
    Can someone help me understand changing between bases?

    For example, if we have

    (34)_5 + (23)_5, I think of it this way:

    We have 5 5's and then 7 1's so we have 6 5's and 2 1's which means we have:

    1 (5^2) .. 1 (5^1) .. 2 (5^0)

    So our answer then would be (112)_5?
    Yes !


    Alternatively, can we just say (34)_5 + (23)_5 = (57)_5? I'm assuming not, since we're using base 10 rules in adding.
    You can't say so because since you're in base 5, the numbers' digits can only be represented by 0,1,2,3 or 4.
    So basically, yes, it's not the same rules as base 10.
    (4+3)_5=(5*1)_5+2=(12)_5

    Instead, I think it would be (34)_5 + (23)_5 = (62)_5
    No ! Because it's the same problem for 6.
    (6)_5=(5*1)_5+1=11.


    Now, if you want to check :

    (34)_5=3 \times 5^1+4 \times 5^0

    (23)_5=2 \times 5^1+3 \times 5^0

    Thus (34)_5+(23)_5=5 \times 5^1+7 \times 5^0=(1 \times 5^2)+(5 \times 5^0+2 \times 5^0)={\color{red}1} \times 5^2+{\color{red}1} \times 5^1+{\color{red}2} \times 5^0=({\color{red}112})_5


    Does this look clear to you ?
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  5. #5
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    Quote Originally Posted by DiscreteW View Post
    Can someone help me understand changing between bases?

    For example, if we have

    (34)_5 + (23)_5, I think of it this way:

    We have 5 5's and then 7 1's so we have 6 5's and 2 1's which means we have:

    1 (5^2) .. 1 (5^1) .. 2 (5^0)

    So our answer then would be (112)_5?

    Alternatively, can we just say (34)_5 + (23)_5 = (57)_5? I'm assuming not, since we're using base 10 rules in adding.

    Instead, I think it would be (34)_5 + (23)_5 = (62)_5

    Some advice?
    (34)_5 + (23)_5

    means:

    (3 \times 5+4 )+(2 \times 5 + 3)

    in base 10:

    (3 \times 5+4 )+(2 \times 5 + 3)=5 \times 5 + 7 = 1 \times 25 + 1 \times 5 +2=(112)_5

    Or you could just compile the base 5 addition table

    RonL
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