SOLUTION: Prove that if 0 < a < b, then 0 < (1/b) < (1/a). Can any of the proofs posted today be proven by using the Direct Proof Method?

Algebra ->  Inequalities -> SOLUTION: Prove that if 0 < a < b, then 0 < (1/b) < (1/a). Can any of the proofs posted today be proven by using the Direct Proof Method?      Log On


   



Question 1207709: Prove that if 0 < a < b, then 0 < (1/b) < (1/a).
Can any of the proofs posted today be proven by using the Direct Proof Method?

Found 2 solutions by Edwin McCravy, math_tutor2020:
Answer by Edwin McCravy(20062) About Me  (Show Source):
You can put this solution on YOUR website!

0%22%22%3C%22%22a%22%22%3C%22%22b

a%22%22%3C%22%22b

0%22%22%3C%22%22b-a

ab%22%22%3E%22%220 since a > 0 and b > 0

0%22%22%3C%22%22%28b-a%29%2F%28ab%29

0%22%22%3C%22%221%2Fa-1%2Fb

1%2Fb%22%22%3C%22%221%2Fa

0%22%22%3C%22%221%2Fb%22%22%3C%22%221%2Fa since 1/b > 0

Edwin

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

0 < a < b
a < b
a*(1/a) < b*(1/a)
1 < b/a
1*(1/b) < (b/a)*(1/b)
1/b < 1/a
0 < 1/b < 1/a

Since a > 0, we know that 1/a > 0 also.
Multiplying both sides by a positive number will not flip the inequality sign.
The same applies for b as well.
b > 0 leads to 1/b > 0.

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Another approach

0 < a < b
a < b
a*(1/(ab)) < b*(1/(ab))
1/b < 1/a
0 < 1/b < 1/a

Note that a > 0 and b > 0 lead to ab > 0, and furthermore 1/(ab) > 0.