Practice Contradiction in Math

Use these practice problems to test your method after reviewing the concept explanation and worked examples.

Worksheet PDF Free Answer-key PDF Family

Quick Recap

A mathematical statement that is always false — no values of the variables can ever make it true.

$x + y = 5$ AND $x + y = 7$ can't both be true simultaneously — this is a contradiction.

Showing a random 20 of 50 problems.

Example 1

easy

0 = 0

a contradiction or an identity?

Example 2

medium

Elimination on a

3\times3

system yields

0 = 1

in one row. What can you conclude?

Example 3

medium

Solve

\dfrac{2}{x-1} = \dfrac{2}{x-1} + 5

Example 4

easy

Does reaching

0 = 3

mean you made an arithmetic error?

Example 5

easy

0 = 5

ever true?

Example 6

easy

True or false:

\sqrt{x} = -2

has no real solution.

Example 7

easy

Does

x^2 = -9

have a real solution?

Example 8

hard

Prove by contradiction that there is no smallest positive real number.

Example 9

medium

Does

\{2x = 6,\ 2x = 10\}

contain a contradiction?

Example 10

medium

Does the system

\{x + y > 5,\ x + y < 2\}

have a solution?

Example 11

medium

Solve

|x - 3| = -1

Example 12

easy

How many solutions does

3(x+2) = 3x + 6

have?

Example 13

medium

What is the difference between a contradiction and an identity in elementary algebra?

Example 14

medium

After elimination you get

0 = 0

. Is this a contradiction?

Example 15

medium

Solve

2(x + 3) = 2x + 5

Example 16

easy

System reduces to

0 = -2

. How many solutions?

Example 17

medium

Solve

5(x-2) - 3x = 2x + 7

Example 18

medium

If a derivation under your assumption produces a contradiction, what can you conclude about the assumption?

Example 19

medium

Does the system

\{y = 2x + 1,\ y = 2x - 3\}

have a solution?

Example 20

easy

Can

x + 3 = x + 7

have a solution?

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Related Concepts

Equations Consistency