How do you simplify #(-6)^0#?

3 Answers
Mar 20, 2016

Answer:

#color(blue)(=1#

Explanation:

  • As per property:
    #color(blue)(a^0= 1#

So, applying the above:

#(-6)^0 =1#

Mar 20, 2016

Answer:

1

Explanation:

#(-6)^0=(-6)^(n-n)=(-6)^n/(-6)^n=1#

Mar 20, 2016

Answer:

#(-6)^0 = 1#

Explanation:

If #n# is a positive integer, then for any number #a# we can define:

#a^n = overbrace(a xx a xx .. xx a)^"n times"#

This has the pleasing property that if #m# and #n# are positive integers then:

#a^m * a^n = overbrace(a xx a xx .. xx a)^"m times" xx overbrace(a xx a xx .. xx a)^"n times"#

#=overbrace(a xx a xx .. xx a)^"m + n times" = a^(m+n)#

If #a != 0# then we can also define #a^(-n) = 1/underbrace(a xx a xx .. xx a)_"n times"#

With this definition we find that #a#'s in numerators and denominators combine or cancel in such a way that:

#a^m * a^n = a^(m+n)#

for any integers #m# and #n# (positive or negative).

In particular. If #a != 0# then #a^0 = a^(1-1) = a^1 * a^(-1) = a/a = 1#

In fact, in the case #n=0# we can think of #a^0# as being an empty product of #0# copies of #a#. The empty product has the value #1# - the identity under multiplication.

#color(white)()#
In particular, if #a = -6# then #a^0 = 1#