Question

To show:a < frac{a+b}{2} < bGiven information:a and b are real numbers.

Transformation properties
ANSWERED
asked 2021-02-01

To show:
\(\displaystyle{a}\ {<}\ {\frac{{{a}+{b}}}{{{2}}}}\ {<}\ {b}\)
Given information:
a and b are real numbers.

Answers (1)

2021-02-02

Concept used:
Number which can be written on number line called real number.
Calculation:
\(\displaystyle\text{Let}\ {a}={2}\ \text{and}\ {b}={3}\)
\(\displaystyle{a}\ {<}\ {b}\)
\(\displaystyle{2}\ {<}\ {3}\)
Then
\(\displaystyle{\frac{{{a}\ +\ {b}}}{{{2}}}}={\frac{{{2}\ +\ {3}}}{{{2}}}}\)
\(\displaystyle={\frac{{{5}}}{{{2}}}}\)
\(\displaystyle={2.5}\)
Hence, \(\displaystyle{\frac{{{a}\ +\ {b}}}{{{2}}}}\) is in between a and b.
\(\displaystyle{a}\ {<}\ {\frac{{{a}\ +\ {b}}}{{{2}}}}\ {<}\ {b}\)
Conclusion:
For real number a and b, if \(\displaystyle{a}\ {<}\ {b}\ \text{then}{<}\ {\frac{{{a}\ +\ {b}}}{{{2}}}}\ {<}\ {b}\)

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