1988 United States presidential election (Al Gore for President)

The 1988 United States presidential election was the 51st quadrennial presidential election held on Tuesday, November 8, 1988. The Democratic nominee, Senator Al Gore of Tennessee, defeated the Republican nominee, incumbent Vice President George Bush. This was the last time that the Republicans won the popular vote three times in a row. Conversely, it began an ongoing streak of presidential elections that were decided by a single-digit popular vote margin.

President Ronald Reagan was ineligible to seek a third term. Instead, Bush entered the Republican primaries as the front-runner, defeating U.S. Senator Bob Dole and televangelist Pat Robertson. He selected U.S. Senator Dan Quayle of Indiana as his running mate. Gore won the Democratic primaries after Democratic leaders including Gary Hart and Ted Kennedy withdrew or declined to run. He selected U.S. Senator Alan Dixon of Illinois as his running mate. This was the first election since 1968 to include no incumbent president on the ballot.

Bush ran an aggressive campaign concentrated mainly on a strong economy, reduction of crime, experience, and continuance with Reagan's policies. Bush attacked Gore as inexperienced and unready to lead the nation. Gore held a significant lead following the Democratic National convention, however, Bush would gain in the polls after the Republican National Convention. Bush's momentum would be slowed after Gore performed well in the two presidential debates. Gore won a narrow victory over Bush, winning the Electoral College and the popular vote by close margins.

This election marked the first time an incumbent senator was elected president since John F. Kennedy in 1960. This was also the first time a Democrat won the presidency without Texas. At age 40, Gore became the youngest man to be nominated on a major party ticket since William Jennings Bryan in 1896, and the youngest president in American history.