-0.000 000 000 742 132 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 132 7(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
-0.000 000 000 742 132 7(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:

|-0.000 000 000 742 132 7| = 0.000 000 000 742 132 7


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: 0.000 000 000 742 132 7.

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) 0.000 000 000 742 132 7 × 2 = 0 + 0.000 000 001 484 265 4;
  • 2) 0.000 000 001 484 265 4 × 2 = 0 + 0.000 000 002 968 530 8;
  • 3) 0.000 000 002 968 530 8 × 2 = 0 + 0.000 000 005 937 061 6;
  • 4) 0.000 000 005 937 061 6 × 2 = 0 + 0.000 000 011 874 123 2;
  • 5) 0.000 000 011 874 123 2 × 2 = 0 + 0.000 000 023 748 246 4;
  • 6) 0.000 000 023 748 246 4 × 2 = 0 + 0.000 000 047 496 492 8;
  • 7) 0.000 000 047 496 492 8 × 2 = 0 + 0.000 000 094 992 985 6;
  • 8) 0.000 000 094 992 985 6 × 2 = 0 + 0.000 000 189 985 971 2;
  • 9) 0.000 000 189 985 971 2 × 2 = 0 + 0.000 000 379 971 942 4;
  • 10) 0.000 000 379 971 942 4 × 2 = 0 + 0.000 000 759 943 884 8;
  • 11) 0.000 000 759 943 884 8 × 2 = 0 + 0.000 001 519 887 769 6;
  • 12) 0.000 001 519 887 769 6 × 2 = 0 + 0.000 003 039 775 539 2;
  • 13) 0.000 003 039 775 539 2 × 2 = 0 + 0.000 006 079 551 078 4;
  • 14) 0.000 006 079 551 078 4 × 2 = 0 + 0.000 012 159 102 156 8;
  • 15) 0.000 012 159 102 156 8 × 2 = 0 + 0.000 024 318 204 313 6;
  • 16) 0.000 024 318 204 313 6 × 2 = 0 + 0.000 048 636 408 627 2;
  • 17) 0.000 048 636 408 627 2 × 2 = 0 + 0.000 097 272 817 254 4;
  • 18) 0.000 097 272 817 254 4 × 2 = 0 + 0.000 194 545 634 508 8;
  • 19) 0.000 194 545 634 508 8 × 2 = 0 + 0.000 389 091 269 017 6;
  • 20) 0.000 389 091 269 017 6 × 2 = 0 + 0.000 778 182 538 035 2;
  • 21) 0.000 778 182 538 035 2 × 2 = 0 + 0.001 556 365 076 070 4;
  • 22) 0.001 556 365 076 070 4 × 2 = 0 + 0.003 112 730 152 140 8;
  • 23) 0.003 112 730 152 140 8 × 2 = 0 + 0.006 225 460 304 281 6;
  • 24) 0.006 225 460 304 281 6 × 2 = 0 + 0.012 450 920 608 563 2;
  • 25) 0.012 450 920 608 563 2 × 2 = 0 + 0.024 901 841 217 126 4;
  • 26) 0.024 901 841 217 126 4 × 2 = 0 + 0.049 803 682 434 252 8;
  • 27) 0.049 803 682 434 252 8 × 2 = 0 + 0.099 607 364 868 505 6;
  • 28) 0.099 607 364 868 505 6 × 2 = 0 + 0.199 214 729 737 011 2;
  • 29) 0.199 214 729 737 011 2 × 2 = 0 + 0.398 429 459 474 022 4;
  • 30) 0.398 429 459 474 022 4 × 2 = 0 + 0.796 858 918 948 044 8;
  • 31) 0.796 858 918 948 044 8 × 2 = 1 + 0.593 717 837 896 089 6;
  • 32) 0.593 717 837 896 089 6 × 2 = 1 + 0.187 435 675 792 179 2;
  • 33) 0.187 435 675 792 179 2 × 2 = 0 + 0.374 871 351 584 358 4;
  • 34) 0.374 871 351 584 358 4 × 2 = 0 + 0.749 742 703 168 716 8;
  • 35) 0.749 742 703 168 716 8 × 2 = 1 + 0.499 485 406 337 433 6;
  • 36) 0.499 485 406 337 433 6 × 2 = 0 + 0.998 970 812 674 867 2;
  • 37) 0.998 970 812 674 867 2 × 2 = 1 + 0.997 941 625 349 734 4;
  • 38) 0.997 941 625 349 734 4 × 2 = 1 + 0.995 883 250 699 468 8;
  • 39) 0.995 883 250 699 468 8 × 2 = 1 + 0.991 766 501 398 937 6;
  • 40) 0.991 766 501 398 937 6 × 2 = 1 + 0.983 533 002 797 875 2;
  • 41) 0.983 533 002 797 875 2 × 2 = 1 + 0.967 066 005 595 750 4;
  • 42) 0.967 066 005 595 750 4 × 2 = 1 + 0.934 132 011 191 500 8;
  • 43) 0.934 132 011 191 500 8 × 2 = 1 + 0.868 264 022 383 001 6;
  • 44) 0.868 264 022 383 001 6 × 2 = 1 + 0.736 528 044 766 003 2;
  • 45) 0.736 528 044 766 003 2 × 2 = 1 + 0.473 056 089 532 006 4;
  • 46) 0.473 056 089 532 006 4 × 2 = 0 + 0.946 112 179 064 012 8;
  • 47) 0.946 112 179 064 012 8 × 2 = 1 + 0.892 224 358 128 025 6;
  • 48) 0.892 224 358 128 025 6 × 2 = 1 + 0.784 448 716 256 051 2;
  • 49) 0.784 448 716 256 051 2 × 2 = 1 + 0.568 897 432 512 102 4;
  • 50) 0.568 897 432 512 102 4 × 2 = 1 + 0.137 794 865 024 204 8;
  • 51) 0.137 794 865 024 204 8 × 2 = 0 + 0.275 589 730 048 409 6;
  • 52) 0.275 589 730 048 409 6 × 2 = 0 + 0.551 179 460 096 819 2;
  • 53) 0.551 179 460 096 819 2 × 2 = 1 + 0.102 358 920 193 638 4;
  • 54) 0.102 358 920 193 638 4 × 2 = 0 + 0.204 717 840 387 276 8;

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:


0.000 000 000 742 132 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1011 1100 10(2)

6. Positive number before normalization:

0.000 000 000 742 132 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1011 1100 10(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:


0.000 000 000 742 132 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1011 1100 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1011 1100 10(2) × 20 =

1.1001 0111 1111 1101 1110 010(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 1101 1110 010


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 1110 1111 0010 =

100 1011 1111 1110 1111 0010


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 1110 1111 0010


Decimal number -0.000 000 000 742 132 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1110 1111 0010

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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