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Suppose that $a$ and $b-1$ are consecutive even numbers and $a<b-1$. It seems that $3a$ and $3b-8$ are consecutive as well, although I am unable to show this.

I realize that $|a-(b-1)| = |a-b+1| = 2$ and

$$|3a-(3b-8)| = |3(a-b)+8| = |3(a-b)+3+5| = |3(a-b+1)+5|$$

If we use $a-b+1 = \pm 2$,

$$|3a-(3b-8)| = |\pm6+5|$$

So I do not see how they can be consecutive, because their difference won't be $2$. But I'm probably not on the right track. Can I hear your feedback?

EDIT: Their difference is $1$ and not $2$, so for the case that $a-b+1 = -2$ which means $a<b-1$, one obtains

$$|3a-(3b-8)| = 1$$

In which case they are consecutive.

  • This is not correct. Take $a=4,b=3$. Then $a$ and $b-1$ are consecutive even integers, but $3a$ and $3b-8$ are not consecutive integers. – ShyamalSayak Jan 13 '24 at 17:34
  • @ShyamalSayak Yes, which is why we need to assume that $a<b-1$ as I wrote in the recent edit. –  Jan 13 '24 at 17:45

2 Answers2

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$a$ can't equal $b$ as then $a$ and $b-1$ won't be both even.

Suppose $a<b$ and $a$ and $b-1$ are consecutive even numbers. Then, $a=2k$ and $b-1=2k+2$ for some whole number $k.$ Then, $3a=6k.$ Moreover, $3b-8=3(2k+3)-8=6k+1.$ Then, $3a$ and $3b-8$ are consecutive.

Suppose $a>b.$ Then, $a=2m$ and $b-1=2m-2$ for some natural number $m.$ Then, $3a=6m.$ Moreover, $3b-8=3(2m-1)-8=6m-11.$ In this case, they won't be consecutive.

Why are you checking to see if their difference is $2?$ Their difference must be $1$ for them to be consecutive.

aqualubix
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  • How do you know that $a<b-1$? Well if you take two even numbers, $a,b-1$, then their difference must be $2$. Because if they differ by $1$, then one of them is odd which is not true. –  Jan 13 '24 at 17:09
  • @Keane, check my edit. – aqualubix Jan 13 '24 at 17:14
  • I thought you wanted to show that $3a$ and $3b-8$ will be consecutive numbers (not consecutive even numbers). It is not possible for them to be consecutive even numbers. – aqualubix Jan 13 '24 at 17:16
  • Ahh, you're right. Please do not mind my lack of attention :) –  Jan 13 '24 at 17:17
  • @Keane, that's alright :) – aqualubix Jan 13 '24 at 17:18
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If $a$ and $b-1$ are consecutive even numbers, then

$$ a=2k \\ b-1=2k+2\\ b=2k+3 $$

So,

$$ 3a=6k\\ 3b-8=3(2k+3)-8\\ 3b-8=6k+9-8\\ 3b-8=6k+1\\ 3b-8=3a+1 $$

Therefore $3a$ and $3b-8$ are consecutive.

Om Patil
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