Alex Rodriguez to retire

Three-time American League MVP and 14-time All-Star Alex Rodriguez is set to play his final game on Aug. 12, the Yankees announced on Sunday prior to a press conference at Yankee Stadium.

Rodriguez will be unconditionally released from his player contract after the game and join the Yankees as a special adviser and instructor, ending a 22-year MLB career. That adviser-instructor deal will last until Dec. 31, 2017, the Yankees announced.

"This is a tough day," an emotional Rodriguez said. "I love this game and I love this team, and today I'm saying goodbye to both."

Rodriguez's 696 career home runs is fourth-all time, behind only Barry Bonds, Hank Aaron and Babe Ruth.
Rodriguez's announcement on Sunday came just two days after teammate Mark Teixeira announced he would retire at the end of the 2016 season.

"Saying 'Goodbye' may be the hardest part of the job, but that's what I'm doing today," Rodriguez said.

Rodriguez, who turned 41 on July 27, has hit .204 with nine home runs in 62 games this season. Since July 16, Rodriguez is 2-for-26 with one home run. The down year comes after a strong 2015 in which he hit 33 home runs.

The first overall pick by the Mariners in the 1993 draft, Rodriguez made his MLB debut the next season. A perennial All-Star at the shortstop position, Rodriguez signed a 10-year, $252 million contract with the Rangers prior to the 2001 season. But the team, mired at the bottom of the AL West, traded Rodriguez to the Yankees in a blockbuster deal before the 2004 season.

To accommodate incumbent Yankees shortstop Derek Jeter, Rodriguez switched to third base upon his arrival in New York. After his MVP 2007 season with the Yankees, Rodriguez opted out of his original contract and signed a new 10-year, $275 million contract with multiple bonus clauses. Both his original Texas contract and the 2007 Yankees deal were, at the time of their signing, the biggest contracts ever signed by an MLB player.