-0.000 000 000 742 146 37 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 146 37(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 742 146 37(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 742 146 37| = 0.000 000 000 742 146 37


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 742 146 37.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 742 146 37 × 2 = 0 + 0.000 000 001 484 292 74;
  • 2) 0.000 000 001 484 292 74 × 2 = 0 + 0.000 000 002 968 585 48;
  • 3) 0.000 000 002 968 585 48 × 2 = 0 + 0.000 000 005 937 170 96;
  • 4) 0.000 000 005 937 170 96 × 2 = 0 + 0.000 000 011 874 341 92;
  • 5) 0.000 000 011 874 341 92 × 2 = 0 + 0.000 000 023 748 683 84;
  • 6) 0.000 000 023 748 683 84 × 2 = 0 + 0.000 000 047 497 367 68;
  • 7) 0.000 000 047 497 367 68 × 2 = 0 + 0.000 000 094 994 735 36;
  • 8) 0.000 000 094 994 735 36 × 2 = 0 + 0.000 000 189 989 470 72;
  • 9) 0.000 000 189 989 470 72 × 2 = 0 + 0.000 000 379 978 941 44;
  • 10) 0.000 000 379 978 941 44 × 2 = 0 + 0.000 000 759 957 882 88;
  • 11) 0.000 000 759 957 882 88 × 2 = 0 + 0.000 001 519 915 765 76;
  • 12) 0.000 001 519 915 765 76 × 2 = 0 + 0.000 003 039 831 531 52;
  • 13) 0.000 003 039 831 531 52 × 2 = 0 + 0.000 006 079 663 063 04;
  • 14) 0.000 006 079 663 063 04 × 2 = 0 + 0.000 012 159 326 126 08;
  • 15) 0.000 012 159 326 126 08 × 2 = 0 + 0.000 024 318 652 252 16;
  • 16) 0.000 024 318 652 252 16 × 2 = 0 + 0.000 048 637 304 504 32;
  • 17) 0.000 048 637 304 504 32 × 2 = 0 + 0.000 097 274 609 008 64;
  • 18) 0.000 097 274 609 008 64 × 2 = 0 + 0.000 194 549 218 017 28;
  • 19) 0.000 194 549 218 017 28 × 2 = 0 + 0.000 389 098 436 034 56;
  • 20) 0.000 389 098 436 034 56 × 2 = 0 + 0.000 778 196 872 069 12;
  • 21) 0.000 778 196 872 069 12 × 2 = 0 + 0.001 556 393 744 138 24;
  • 22) 0.001 556 393 744 138 24 × 2 = 0 + 0.003 112 787 488 276 48;
  • 23) 0.003 112 787 488 276 48 × 2 = 0 + 0.006 225 574 976 552 96;
  • 24) 0.006 225 574 976 552 96 × 2 = 0 + 0.012 451 149 953 105 92;
  • 25) 0.012 451 149 953 105 92 × 2 = 0 + 0.024 902 299 906 211 84;
  • 26) 0.024 902 299 906 211 84 × 2 = 0 + 0.049 804 599 812 423 68;
  • 27) 0.049 804 599 812 423 68 × 2 = 0 + 0.099 609 199 624 847 36;
  • 28) 0.099 609 199 624 847 36 × 2 = 0 + 0.199 218 399 249 694 72;
  • 29) 0.199 218 399 249 694 72 × 2 = 0 + 0.398 436 798 499 389 44;
  • 30) 0.398 436 798 499 389 44 × 2 = 0 + 0.796 873 596 998 778 88;
  • 31) 0.796 873 596 998 778 88 × 2 = 1 + 0.593 747 193 997 557 76;
  • 32) 0.593 747 193 997 557 76 × 2 = 1 + 0.187 494 387 995 115 52;
  • 33) 0.187 494 387 995 115 52 × 2 = 0 + 0.374 988 775 990 231 04;
  • 34) 0.374 988 775 990 231 04 × 2 = 0 + 0.749 977 551 980 462 08;
  • 35) 0.749 977 551 980 462 08 × 2 = 1 + 0.499 955 103 960 924 16;
  • 36) 0.499 955 103 960 924 16 × 2 = 0 + 0.999 910 207 921 848 32;
  • 37) 0.999 910 207 921 848 32 × 2 = 1 + 0.999 820 415 843 696 64;
  • 38) 0.999 820 415 843 696 64 × 2 = 1 + 0.999 640 831 687 393 28;
  • 39) 0.999 640 831 687 393 28 × 2 = 1 + 0.999 281 663 374 786 56;
  • 40) 0.999 281 663 374 786 56 × 2 = 1 + 0.998 563 326 749 573 12;
  • 41) 0.998 563 326 749 573 12 × 2 = 1 + 0.997 126 653 499 146 24;
  • 42) 0.997 126 653 499 146 24 × 2 = 1 + 0.994 253 306 998 292 48;
  • 43) 0.994 253 306 998 292 48 × 2 = 1 + 0.988 506 613 996 584 96;
  • 44) 0.988 506 613 996 584 96 × 2 = 1 + 0.977 013 227 993 169 92;
  • 45) 0.977 013 227 993 169 92 × 2 = 1 + 0.954 026 455 986 339 84;
  • 46) 0.954 026 455 986 339 84 × 2 = 1 + 0.908 052 911 972 679 68;
  • 47) 0.908 052 911 972 679 68 × 2 = 1 + 0.816 105 823 945 359 36;
  • 48) 0.816 105 823 945 359 36 × 2 = 1 + 0.632 211 647 890 718 72;
  • 49) 0.632 211 647 890 718 72 × 2 = 1 + 0.264 423 295 781 437 44;
  • 50) 0.264 423 295 781 437 44 × 2 = 0 + 0.528 846 591 562 874 88;
  • 51) 0.528 846 591 562 874 88 × 2 = 1 + 0.057 693 183 125 749 76;
  • 52) 0.057 693 183 125 749 76 × 2 = 0 + 0.115 386 366 251 499 52;
  • 53) 0.115 386 366 251 499 52 × 2 = 0 + 0.230 772 732 502 999 04;
  • 54) 0.230 772 732 502 999 04 × 2 = 0 + 0.461 545 465 005 998 08;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 742 146 37(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1010 00(2)

6. Positive number before normalization:

0.000 000 000 742 146 37(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1010 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 146 37(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1010 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1010 00(2) × 20 =

1.1001 0111 1111 1111 1101 000(2) × 2-31


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1001 0111 1111 1111 1101 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1011 1111 1111 1110 1000 =

100 1011 1111 1111 1110 1000


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 1111 1110 1000


Decimal number -0.000 000 000 742 146 37 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1110 1000

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111