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I'm looking for a contradiction or a proof.

Suppose we have the following inequality:

$a\cdot j \leq b \cdot h$

Where $a,b,j,h \in \mathbb{N}$.

We now replace $a$ with $a_{1}$ and $b$ with $b_{1}$, where $a_{1}, b_{1} \in \mathbb{N}$. We also have the following situation:

  1. $a_{1}\leq a$
  2. $b_{1}\leq b$
  3. $b-b_{1}\leq a-a_{1}$

Can we then deduce that $a_{1}\cdot j \leq b_{1} \cdot h$ is true?

For instance:

$4\cdot 6 \leq 7 \cdot 5$

In this case $a = 4, b=7$. We now replace $a$ with $a_{1} = 3$ so $a - a_{1} = 1$. Since $b-b_{1}\leq a-a_{1} = 1$ we have $b_{1}\in\lbrace 6,7 \rbrace$. In both cases:

  1. $3\cdot 6 \leq 7 \cdot 5$
  2. $3\cdot 6 \leq 6 \cdot 5$

The inequality with the replacements of $a,b$ is true.

user11775
  • 333

2 Answers2

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So, will this work?
Initially: $$b-b_{1}\leq a-a_{1}.$$ Multiply by $h$: $$b \ h-b_{1} h\leq a \ h-a_{1} h.$$ Use the first inequality: $$a\ j - b_1 h\leq b\ h-b_{1} h\leq a \ h-a_{1} h$$ $$a\ j - b_1 h\leq a \ h-a_{1} h.$$ Rearrange the terms: $$a \ (j-h)+a_1 h \leq b_1 h.$$ Use equation $(1)$: $$a_1(j-h)+a_1 h\leq a \ (j-h)+a_1 h \leq b_1 h.$$ And get: $$a_1j\leq b_1 h$$ Well, but you'll need $(j-h)\geq 0$.

Caran-d'Ache
  • 3,564
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We cannot, since it is false in general: let $j=1,h=5,a=10,b=2$, so that $$ a\cdot j=10\cdot 1=10 \leq 10=2\cdot 5=b\cdot h $$ then let $a_1=9,b_1=1$, thus

  1. $a_1=9\leq 10=a$
  2. $b_1=1\leq 2=b$
  3. $b-b_1=2-1=1\leq 1=10-9=a-a_1$

and $$ a_1\cdot j =9\cdot 1=9 \geq 5=1\cdot 5=b_1\cdot h $$

A.P.
  • 9,728