Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Elementary algebra
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
=== Substitution === {{main|Substitution (algebra)}} {{see also|Substitution (logic)}} Substitution is replacing the terms in an expression to create a new expression. Substituting 3 for {{mvar|a}} in the expression {{math|''a''*5}} makes a new expression {{math|3*5}} with meaning {{math|15}}. Substituting the terms of a statement makes a new statement. When the original statement is true independently of the values of the terms, the statement created by substitutions is also true. Hence, definitions can be made in symbolic terms and interpreted through substitution: if <math>a^2:=a\times a</math> is meant as the definition of <math>a^2,</math> as the product of {{mvar|a}} with itself, substituting {{math|3}} for {{mvar|a}} informs the reader of this statement that <math>3^2</math> means {{math|1=3 Γ 3 = 9}}. Often it's not known whether the statement is true independently of the values of the terms. And, substitution allows one to derive restrictions on the possible values, or show what conditions the statement holds under. For example, taking the statement {{math|1=''x'' + 1 = 0}}, if {{mvar|x}} is substituted with {{math|1}}, this implies {{math|1=1 + 1 = 2 = 0}}, which is false, which implies that if {{math|1=''x'' + 1 = 0}} then {{mvar|x}} cannot be {{math|1}}. If {{math|''x''}} and {{math|''y''}} are [[integers]], [[rationals]], or [[real numbers]], then {{math|1=''xy'' = 0}} implies {{math|1=''x'' = 0}} or {{math|1=''y'' = 0}}. Consider {{math|1=''abc'' = 0}}. Then, substituting {{math|''a''}} for {{math|''x''}} and {{math|''bc''}} for {{math|''y''}}, we learn {{math|1=''a'' = 0}} or {{math|1=''bc'' = 0}}. Then we can substitute again, letting {{math|1=''x'' = ''b''}} and {{math|1=''y'' = ''c''}}, to show that if {{math|1=''bc'' = 0}} then {{math|1=''b'' = 0}} or {{math|1=''c'' = 0}}. Therefore, if {{math|1=''abc'' = 0}}, then {{math|1=''a'' = 0}} or ({{math|1=''b'' = 0}} or {{math|1=''c'' = 0}}), so {{math|1=''abc'' = 0}} implies {{math|1=''a'' = 0}} or {{math|1=''b'' = 0}} or {{math|1=''c'' = 0}}. If the original fact were stated as "{{math|1=''ab'' = 0}} implies {{math|1=''a'' = 0}} or {{math|1=''b'' = 0}}", then when saying "consider {{math|1=''abc'' = 0}}," we would have a conflict of terms when substituting. Yet the above logic is still valid to show that if {{math|1=''abc'' = 0}} then {{math|1=''a'' = 0}} or {{math|1=''b'' = 0}} or {{math|1=''c'' = 0}} if, instead of letting {{math|1=''a'' = ''a''}} and {{math|1=''b'' = ''bc''}}, one substitutes {{math|''a''}} for {{math|''a''}} and {{math|''b''}} for {{math|''bc''}} (and with {{math|1=''bc'' = 0}}, substituting {{math|''b''}} for {{math|''a''}} and {{math|''c''}} for {{math|''b''}}). This shows that substituting for the terms in a statement isn't always the same as letting the terms from the statement equal the substituted terms. In this situation it's clear that if we substitute an expression {{math|''a''}} into the {{math|''a''}} term of the original equation, the {{math|''a''}} substituted does not refer to the {{math|''a''}} in the statement "{{math|1=''ab'' = 0}} implies {{math|1=''a'' = 0}} or {{math|1=''b'' = 0}}."
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)