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I need to check if the input argument of a bash script is a base ten number. How can I do?

Your can use pattern matching (improve on the regex if you don't want to accept a leading 0):

if [[ $num =~ ^[+-]?[1-9][0-9]*$ ]]

then

echo Decimal

else

echo Not decimal

fi

Compare it to itself while forcing base-ten interpretation.
As seen in the shell Arithmetic manual:

Constants with a leading 0 are interpreted as octal numbers. A leading
‘0x’ or ‘0X’ denotes hexadecimal. Otherwise, numbers take the form
[base#]n, where the optional base is a decimal number between 2 and 64
representing the arithmetic base, and n is a number in that base. If
base# is omitted, then base 10 is used. When specifying n, the digits
greater than 9 are represented by the lowercase letters, the uppercase
letters, ‘@’, and ‘_’, in that order. If base is less than or equal to
36, lowercase and uppercase letters may be used interchangeably to
represent numbers between 10 and 35.

Therefore if it's the first argument you can do...

EDITED AGAIN TO ACCOMMODATE NEGATIVE INPUT

#!/bin/bash

b=${1#-}

if [[ $b =~ ^[0-9]+$ ]]; then                       #If it's all numbers

        a=$((10#$b))                                #create a base 10 version

        if [ "$a" == "$b" ]; then                   #if they LOOK the same...

                echo "$b is Base 10!"; exit 1; fi;  #It's base 10; exit.

fi

    echo "$b is not base 10"                        #If it has letters or $a and $b look different, it's not base 10.

This code will take any input and tell you if it is base 10 or not. The arg is stripped of a negative sign first, then compared with $a non-numerically, because 012 -eq $((10#012)); but we don't want the script to tell us it's base ten because it isn't.

root@xxxx:~# ./testing.sh -124124

124124 is Base 10!

root@xxxx:~# ./testing.sh 58135

58135 is Base 10!

root@xxxx:~# ./testing.sh aksjflkfj929148

aksjflkfj929148 is not Base 10