# Squaring both sides of equation and inequality?

1. Jul 19, 2015

### ArmanZ

What is a square of a number? A^2=A*A. If A=B squaring both sides will give A^2=B^2. How I think about squaring is we multiply both sides of A=B by A(we could also do this for B) we get A*A=B*A but A=B so this will result in A*A=B*B.
But if we do this for an inequality, A>B, multiplying both sides by A will yield two scenarios(when A>0 and A<0) A*A>B*A and A*A<B*A in both cases A≠B. So is it possible to square an inequality?

2. Jul 19, 2015

### mgkii

You've almost answered the question yourself - stick some numbers in as an example and you're there:

Here's an inequality: 3 > -4

Square both sides : 9 > 16 (which is incorrect!!)

However, if you try the same with two positive numbers then the inequality remains. So you can't define a rule that says you can square an inequality (without potentially changing the nature of the inequality

3. Jul 19, 2015

### tommyxu3

I think it could be, just for two real numbers $a,b>0.$
$a>b>0,$ then $a^2>b^2.$
$a<b<0,$ then $a^2>b^2.$

4. Jul 20, 2015

### ArmanZ

I think my question is not clear. For example 2>4(both are positive).If we square both sides we get 4>16, which would be correct if we did did not consider the following fact: 4>16⇔2*2>4*4 but how could we get 2*2>4*4 from 2>4 if we have to multiply both sides by different numbers( 2 and 4, 2≠4) which is incorrect in principle. Or am I wrong? Because in equation a=b multiplying both sides by a will yield a*a=b*a but a=b so a*a=b*b, which is a^2=b^2. But in inequality it seems to be different. Thanks for your replies guys.

5. Jul 20, 2015

### mgkii

I'm not totally sure I understand your question so apologies I've advance if this isn't what you asked:

For an equality it doesn't matter what you do to each side (as long as it's the same) as by definition the starting point for that operation is the same. So multiple by 4, square it, add 6 and double it, whatever you do will be the same as the starting point is the same.

For an inequality there's an infinity of ways each side can differ, so unless you put bounds on your function as one of the earlier responders noted, you can't make general observations about the function you apply to each side (square, multiply by 12, etc)

6. Jul 20, 2015

### micromass

Staff Emeritus
Consider two positive numbers $a$ and $b$. If $a<b$, then we can multiply both sides with $a$ to get $a^2 < ab$. We can also multiply both sides with $b$ to get $ab<b^2$. Putting both together gives $a^2<ab<b^2$, which gives $a^2<b^2$.

7. Jul 20, 2015

### Staff: Mentor

No.
You're starting from a statement that is not true -- 2 is NOT larger than 4.

8. Jul 22, 2015

### ArmanZ

Sorry, I made a careless mistake once again

9. Jul 22, 2015

### ArmanZ

For everyone who found this discussion, in this comment I was implying 2<4, 4<16⇔2*2<4*4, 2*2<4*4. But anyway, I hope the general idea of this discussion is understood correctly by everyone.