Dgdrxrt

If x and y are integers, can I rewrite x < y + 1 as x ≤ y?

If x = 1 and y = 1,
x < y + 1 ≡ 1 < 1 + 1 ≡ 1 < 2, is true, and
x ≤ y ≡ 1 ≤ 1, is also true.

If x = 1 and y = 2,
x < y + 1 ≡ 1 < 2 + 1 ≡ 1 < 3, is true, and
x ≤ y ≡ 1 ≤ 2, is also true.

If x = 2 and y = 1,
x < y + 1 ≡ 2 < 1 + 1 ≡ 2 < 2, is false, and
x ≤ y ≡ 2 ≤ 1, is also false.

I don't see why x < y + 1 can't be rewritten as x ≤ y, or the other way around, when x and y are both integers.

Let $x$ and $y$ be integers.

If $xle y$ then $x<y+1.$

On the other hand, if $x>y$ then $x ge y+1$.

Thus, $x le y $ if and only if $x<y+1$.

For integers yes you can.

For any $x,yin mathbb R$ we have $y < y+1$ and there are three possibilities for $x$

i) $x ge y+1$. We were told this is not true.

ii) $y < x < y+ 1$. If $y$ is an integer then there are no integers between $y$ and $y+1$. So if $x$ and $y$ are integers then this is impossible.

iii) $x le y$.

So we are told i) is false; and ii) is impossible. That leaves only iii) $x le y$.

Note: You can't do this in general if $x$ and/or $y$ aren't integers.

==== old obtuse answer below ====

Bear with me:

Let's suppose $A$ is a set of numbers.

If $k = max A$, the maximum element (if any) in $A$.

Then if $x in A$ we have to have $x le k$. That's obvious. If $x$ is in $A$ and $k$ is the largest element in $A$ then $x$, an element, is at most, as large as the largest element.

If $A= $ the set of all integers less than $y+1$, an integer. Then $y+1$ and anything larger is not in $A$ because $y+1 not < y+1$ nor would anything larger be. But $y < y+1$ so $y$ is in $A$. So $y = max A$ because $y$ any thing larger is not.

Okay. So $x < y+1$ means $x$ is in $A$ and so $x le max A = y$.

A bit overkill but....