Cleveland Indians are the first MLB team in 2018 to clinch its division

The Cleveland Indians are the first MLB division champion of 2018, and their clincher on Saturday was a perfect reflection of the gap between the Indians and the rest of the American League Central.

The Indians drubbed the Tigers 15-0 on Saturday in Cleveland, widening their lead over the Twins to a whopping 15½ games, the largest divisional advantage in baseball.

Cleveland has outscored its opponents by 154 runs in 2018, winning its third consecutive division title. The other four teams in the AL Central have been outscored by 560 runs and have a collective .401 winning percentage, which translates into a 97-loss pace over 162 games. It’s the worst division in baseball by a wide margin.

The Indians are 43-22 against the AL Central, and 40-43 against everybody else.

The usual suspects have driven the Indians in 2018, with shortstop Francisco Lindor and third baseman Jose Ramirez both legitimate MVP candidates, and pitcher Corey Kluber will be in the mix for the AL Cy Young Award yet again.

On Saturday, Lindor hit his career-best 35th home run, while Ramirez doubled, tripled, singled and walked, scoring three runs and driving in two.

Cleveland added Oliver Perez to its bullpen in June, then traded for Brad Hand and Adam Cimber in July. Now, Andrew Miller is back as well after an injury-plagued summer, helping Indians relievers post a 3.46 ERA since the All-Star break after a woeful 5.28 bullpen ERA in the putative first half.

The Red Sox already have 102 wins — Boston already clinched a playoff spot but not their division, yet — and will very likely host the AL Wild Card winner in the Division Series, which leaves Cleveland playing the AL West winner in the first round, either the Astros or A’s. The Indians lost four of seven games to Houston this season, and lost four of six games to Oakland.

Both AL Division Series will start on Friday, Oct. 5.