Proof that a1m&lt;a1n with m&lt;n,0&lt;a&lt;1 and n,m∈N

Step 1 Rise both side of the inequality to the power m×n to find an&lt;am that is true since so since n&gt;m an=am×an−m&lt;am because an−m&lt;1

Step 1 Set b=a1m and B=a1n, where 0&lt;a&lt;1, and n&gt;m By definition: bm=Bn=a&lt;1 This implies 0&lt;b, B&lt;1 Bn=BmBn−m=bm Since Bn−m&lt;1 it follows that Bm&gt;bm, or (Bb)m&gt;1, which implies Bb&gt;1

Step 1 Rise both side of the inequality to the power m×n to find an&lt;am that is true since so since n&gt;m an=am×an−m&lt;am because an−m&lt;1

Step 1 Set b=a1m and B=a1n, where 0&lt;a&lt;1, and n&gt;m By definition: bm=Bn=a&lt;1 This implies 0&lt;b, B&lt;1 Bn=BmBn−m=bm Since Bn−m&lt;1 it follows that Bm&gt;bm, or (Bb)m&gt;1, which implies Bb&gt;1

"Proof that a1m<a1n with m<n,0<a<1 and n,m∈N" was answered by students like you

veksetz

Answered question

2022-01-13

Proof that

a^{\frac{1}{m}} < a^{\frac{1}{n}}

with

m < n, 0 < a < 1

and

n, m \in N

Answer & Explanation

Pademagk71

Beginner2022-01-14Added 34 answers

Step 1
Rise both side of the inequality to the power

m \times n

to find

a^{n} < a^{m}

that is true since so since

n > m

a^{n} = a^{m} \times a^{n - m} < a^{m}

because

a^{n - m} < 1

sirpsta3u

Beginner2022-01-15Added 42 answers

Step 1
Set

b = a^{\frac{1}{m}}

and

B = a^{\frac{1}{n}}

,
where

0 < a < 1

,
and

n > m

By definition:

b^{m} = B^{n} = a < 1

This implies

0 < b, B < 1

B^{n} = B^{m} B^{n - m} = b^{m}

Since

B^{n - m} < 1

it follows that

B^{m} > b^{m}

, or

{(\frac{B}{b})}^{m} > 1

, which implies

\frac{B}{b} > 1

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