What version of Go are you using (go version)?

Go version is tip.

Does this issue reproduce with the latest release?

yes

What operating system and processor architecture are you using (go env)?

GOARCH="amd64"
GOBIN=""
GOCACHE="/Users/drchase/Library/Caches/go-build"
GOEXE=""
GOFLAGS=""
GOHOSTARCH="amd64"
GOHOSTOS="darwin"
GOOS="darwin"
GOPATH="/Users/drchase/work/gocode"
GOPROXY=""
GORACE=""
GOROOT="/Users/drchase/work/go"
GOTMPDIR=""
GOTOOLDIR="/Users/drchase/work/go/pkg/tool/darwin_amd64"
GCCGO="gccgo"
CC="clang"
CXX="clang++"
CGO_ENABLED="1"
GOMOD="/dev/null"
CGO_CFLAGS="-g -O2"
CGO_CPPFLAGS=""
CGO_CXXFLAGS="-g -O2"
CGO_FFLAGS="-g -O2"
CGO_LDFLAGS="-g -O2"
PKG_CONFIG="pkg-config"
GOGCCFLAGS="-fPIC -m64 -pthread -fno-caret-diagnostics -Qunused-arguments -fmessage-length=0 -fdebug-prefix-map=/var/folders/gr/vvb66dqx6jl6lh8wckfd5p9w0095tn/T/go-build197631297=/tmp/go-build -gno-record-gcc-switches -fno-common"

What did you do?

math.Remainder(-1,1)
https://play.golang.org/p/EZjSc56D0ax

What did you expect to see?

-0

What did you see instead?

0

Documentation for math.Remainder claims IEEE 754:

Remainder returns the IEEE 754 floating-point remainder of x/y.

IEEE 754-1985 (the copy of the standard I could easily find) Section 5.1, second paragraph:

When y != 0 , the remainder r = x REM y is defined regardless of the
rounding mode by the mathematical relation r = x – y × n , where n is the
integer nearest the exact value x/y ; whenever |n – x/y| = ½ , then n is even.
Thus, the remainder is always exact. If r = 0 , its sign shall be that of x.
Precision control (4.3) shall not apply to the remainder operation.

The cause of the bug is these lines in math/remainder.go:

	if x == y {
		return 0
	}

Changing them to read:

	if x == y {
		if !sign {
			return 0
		}
		return -(x-x) // How to say "-0.0" in Go.
	}

fixes the bug, at least on amd64.