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Sagot :
To verify the given properties, we need to check if each of the statements holds true for the given values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]. We have:
[tex]\[ a = \frac{3}{4}, \quad b = \frac{-1}{2}, \quad c = \frac{1}{2} \][/tex]
### Verification of (i): [tex]\(a + b = b + a\)[/tex]
1. Calculate [tex]\(a + b\)[/tex]:
[tex]\[ a + b = \frac{3}{4} + \frac{-1}{2} \][/tex]
To add these fractions, we need a common denominator, which is 4:
[tex]\[ \frac{3}{4} + \frac{-1}{2} = \frac{3}{4} + \frac{-2}{4} = \frac{3 - 2}{4} = \frac{1}{4} \][/tex]
2. Calculate [tex]\(b + a\)[/tex]:
[tex]\[ b + a = \frac{-1}{2} + \frac{3}{4} \][/tex]
We already calculated the same addition just in reverse order:
[tex]\[ \frac{-1}{2} + \frac{3}{4} = \frac{-2}{4} + \frac{3}{4} = \frac{3 - 2}{4} = \frac{1}{4} \][/tex]
Thus, we find that [tex]\(a + b = b + a\)[/tex]:
[tex]\[ \frac{1}{4} = \frac{1}{4} \][/tex]
Therefore, [tex]\(a + b = b + a\)[/tex] holds true.
### Verification of (ii): [tex]\(a + c = c + a\)[/tex]
1. Calculate [tex]\(a + c\)[/tex]:
[tex]\[ a + c = \frac{3}{4} + \frac{1}{2} \][/tex]
To add these fractions, we need a common denominator, which is 4:
[tex]\[ \frac{3}{4} + \frac{1}{2} = \frac{3}{4} + \frac{2}{4} = \frac{3 + 2}{4} = \frac{5}{4} \][/tex]
2. Calculate [tex]\(c + a\)[/tex]:
[tex]\[ c + a = \frac{1}{2} + \frac{3}{4} \][/tex]
We already calculated the same addition just in reverse order:
[tex]\[ \frac{1}{2} + \frac{3}{4} = \frac{2}{4} + \frac{3}{4} = \frac{2 + 3}{4} = \frac{5}{4} \][/tex]
Thus, we find that [tex]\(a + c = c + a\)[/tex]:
[tex]\[ \frac{5}{4} = \frac{5}{4} \][/tex]
Therefore, [tex]\(a + c = c + a\)[/tex] holds true.
### Verification of (iii): [tex]\((a + b) + c = a + (b + c)\)[/tex]
1. Calculate [tex]\((a + b) + c\)[/tex]:
[tex]\[ (a + b) + c = \left( \frac{3}{4} + \frac{-1}{2} \right) + \frac{1}{2} \][/tex]
We already found [tex]\(a + b\)[/tex]:
[tex]\[ a + b = \frac{1}{4} \][/tex]
Now add [tex]\(c\)[/tex] to this result:
[tex]\[ \frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{1 + 2}{4} = \frac{3}{4} \][/tex]
2. Calculate [tex]\(a + (b + c)\)[/tex]:
[tex]\[ a + (b + c) = \frac{3}{4} + \left( \frac{-1}{2} + \frac{1}{2} \right) \][/tex]
Calculate [tex]\(b + c\)[/tex]:
[tex]\[ \frac{-1}{2} + \frac{1}{2} = 0 \][/tex]
Now add [tex]\(a\)[/tex] to this result:
[tex]\[ \frac{3}{4} + 0 = \frac{3}{4} \][/tex]
Thus, we find that [tex]\((a + b) + c = a + (b + c)\)[/tex]:
[tex]\[ \frac{3}{4} = \frac{3}{4} \][/tex]
Therefore, [tex]\((a + b) + c = a + (b + c)\)[/tex] holds true.
In conclusion, all the given properties have been verified:
1. [tex]\(a + b = b + a\)[/tex]
2. [tex]\(a + c = c + a\)[/tex]
3. [tex]\((a + b) + c = a + (b + c)\)[/tex]
All statements are true as verified.
[tex]\[ a = \frac{3}{4}, \quad b = \frac{-1}{2}, \quad c = \frac{1}{2} \][/tex]
### Verification of (i): [tex]\(a + b = b + a\)[/tex]
1. Calculate [tex]\(a + b\)[/tex]:
[tex]\[ a + b = \frac{3}{4} + \frac{-1}{2} \][/tex]
To add these fractions, we need a common denominator, which is 4:
[tex]\[ \frac{3}{4} + \frac{-1}{2} = \frac{3}{4} + \frac{-2}{4} = \frac{3 - 2}{4} = \frac{1}{4} \][/tex]
2. Calculate [tex]\(b + a\)[/tex]:
[tex]\[ b + a = \frac{-1}{2} + \frac{3}{4} \][/tex]
We already calculated the same addition just in reverse order:
[tex]\[ \frac{-1}{2} + \frac{3}{4} = \frac{-2}{4} + \frac{3}{4} = \frac{3 - 2}{4} = \frac{1}{4} \][/tex]
Thus, we find that [tex]\(a + b = b + a\)[/tex]:
[tex]\[ \frac{1}{4} = \frac{1}{4} \][/tex]
Therefore, [tex]\(a + b = b + a\)[/tex] holds true.
### Verification of (ii): [tex]\(a + c = c + a\)[/tex]
1. Calculate [tex]\(a + c\)[/tex]:
[tex]\[ a + c = \frac{3}{4} + \frac{1}{2} \][/tex]
To add these fractions, we need a common denominator, which is 4:
[tex]\[ \frac{3}{4} + \frac{1}{2} = \frac{3}{4} + \frac{2}{4} = \frac{3 + 2}{4} = \frac{5}{4} \][/tex]
2. Calculate [tex]\(c + a\)[/tex]:
[tex]\[ c + a = \frac{1}{2} + \frac{3}{4} \][/tex]
We already calculated the same addition just in reverse order:
[tex]\[ \frac{1}{2} + \frac{3}{4} = \frac{2}{4} + \frac{3}{4} = \frac{2 + 3}{4} = \frac{5}{4} \][/tex]
Thus, we find that [tex]\(a + c = c + a\)[/tex]:
[tex]\[ \frac{5}{4} = \frac{5}{4} \][/tex]
Therefore, [tex]\(a + c = c + a\)[/tex] holds true.
### Verification of (iii): [tex]\((a + b) + c = a + (b + c)\)[/tex]
1. Calculate [tex]\((a + b) + c\)[/tex]:
[tex]\[ (a + b) + c = \left( \frac{3}{4} + \frac{-1}{2} \right) + \frac{1}{2} \][/tex]
We already found [tex]\(a + b\)[/tex]:
[tex]\[ a + b = \frac{1}{4} \][/tex]
Now add [tex]\(c\)[/tex] to this result:
[tex]\[ \frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{1 + 2}{4} = \frac{3}{4} \][/tex]
2. Calculate [tex]\(a + (b + c)\)[/tex]:
[tex]\[ a + (b + c) = \frac{3}{4} + \left( \frac{-1}{2} + \frac{1}{2} \right) \][/tex]
Calculate [tex]\(b + c\)[/tex]:
[tex]\[ \frac{-1}{2} + \frac{1}{2} = 0 \][/tex]
Now add [tex]\(a\)[/tex] to this result:
[tex]\[ \frac{3}{4} + 0 = \frac{3}{4} \][/tex]
Thus, we find that [tex]\((a + b) + c = a + (b + c)\)[/tex]:
[tex]\[ \frac{3}{4} = \frac{3}{4} \][/tex]
Therefore, [tex]\((a + b) + c = a + (b + c)\)[/tex] holds true.
In conclusion, all the given properties have been verified:
1. [tex]\(a + b = b + a\)[/tex]
2. [tex]\(a + c = c + a\)[/tex]
3. [tex]\((a + b) + c = a + (b + c)\)[/tex]
All statements are true as verified.
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