2 + 2 = 5

Algebra Level 2

Below is the "proof" that 2 + 2 = 5. 2 + 2 = 5.

Step 1: 2 + 2 = 4 2 + 2 = 4
Step 2: = 4 9 2 + 9 2 = 4 - {\frac{9}{2}} + {\frac{9}{2}}
Step 3: = ( 4 9 2 ) 2 + 9 2 = \sqrt{(4 - {\frac{9}{2}})^2} + {\frac{9}{2}}
Step 4: = 16 2 × 4 × 9 2 + ( 9 2 ) 2 + 9 2 = \sqrt{16 - 2 \times 4 \times {\frac{9}{2}} + {(\frac{9}{2})^2}} + {\frac{9}{2}}
Step 5: = 20 + ( 9 2 ) 2 + 9 2 = \sqrt{-20 + {(\frac{9}{2})^2}} + {\frac{9}{2}}
Step 6: = 25 45 + ( 9 2 ) 2 + 9 2 = \sqrt{25 - 45 + {(\frac{9}{2})^2}} + {\frac{9}{2}}
Step 7: = 5 2 2 × 5 × 9 2 + ( 9 2 ) 2 + 9 2 = \sqrt{5^2 - 2 \times 5 \times {\frac{9}{2}} + {(\frac{9}{2})^2}} + {\frac{9}{2}}
Step 8: = ( 5 9 2 ) 2 + 9 2 = \sqrt{(5 -{\frac{9}{2}})^2} + {\frac{9}{2}}
Step 9: 2 + 2 = 5 9 2 + 9 2 2 + 2 = 5 - {\frac{9}{2}} + {\frac{9}{2}}
Step 10: 2 + 2 = 5 \boxed{2 + 2 = 5}

In which step there is the first flaw?

Bonus: What alteration can you make in order to correct this flaw?
4 3 7 5 8 6 2

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1 solution

Naren Bhandari
Feb 5, 2018

In step 3 ( 4 9 2 ) 2 = 4 9 2 = 9 2 4 \begin{aligned} \sqrt{\left(4-\frac{9}{2}\right)^2} = \big| 4-\frac{9}{2}\big | = \frac{9}{2} -4 \end{aligned}

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