Golang Tour Excercise Square Root


I’m trying to figure out why 1e-10 is used in this code. I was running through the exercises and got stuck on Next, change the loop condition to stop once the value has stopped changing (or only changes by a very small delta). See if that's more or fewer iterations. After not figuring it out, I searched and found this solution by someone else.

In a lot of solutions, I saw 1e-10 as part of an if statement within the for loop. In Golang, does it mean that 1e is risen to the (-10) power? Is it like, automatic for Golang? So basically normally I’d assume it would be written as 1e^(-10)?

package main

import (

func Sqrt(x float64) float64 {
    z := float64(2.)
    s := float64(0)
    for i := 0; i < 10; i++ {
        z = z - (z*z - x) / (2 * z)
        if math.Abs(z-s) < 1e-10 {
        s = z
    return z

func main() {

Thanks for any/all help one may provide!


1e-10 is a floating-point literal with the value 1 to the power minus 10.

The Go Programming Language Specification

Floating-point literals

A floating-point literal is a decimal representation of a
floating-point constant. It has an integer part, a decimal point, a
fractional part, and an exponent part. The integer and fractional part
comprise decimal digits; the exponent part is an e or E followed by an
optionally signed decimal exponent. One of the integer part or the
fractional part may be elided; one of the decimal point or the
exponent may be elided.

float_lit = decimals "." [ decimals ] [ exponent ] |
            decimals exponent |
            "." decimals [ exponent ] .
decimals  = decimal_digit { decimal_digit } .
exponent  = ( "e" | "E" ) [ "+" | "-" ] decimals .

072.40  // == 72.40

Answered By – peterSO

Answer Checked By – David Marino (GoLangFix Volunteer)

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