-0.000 000 000 742 147 673 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 673 4(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
-0.000 000 000 742 147 673 4(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:

|-0.000 000 000 742 147 673 4| = 0.000 000 000 742 147 673 4


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: 0.000 000 000 742 147 673 4.

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) 0.000 000 000 742 147 673 4 × 2 = 0 + 0.000 000 001 484 295 346 8;
  • 2) 0.000 000 001 484 295 346 8 × 2 = 0 + 0.000 000 002 968 590 693 6;
  • 3) 0.000 000 002 968 590 693 6 × 2 = 0 + 0.000 000 005 937 181 387 2;
  • 4) 0.000 000 005 937 181 387 2 × 2 = 0 + 0.000 000 011 874 362 774 4;
  • 5) 0.000 000 011 874 362 774 4 × 2 = 0 + 0.000 000 023 748 725 548 8;
  • 6) 0.000 000 023 748 725 548 8 × 2 = 0 + 0.000 000 047 497 451 097 6;
  • 7) 0.000 000 047 497 451 097 6 × 2 = 0 + 0.000 000 094 994 902 195 2;
  • 8) 0.000 000 094 994 902 195 2 × 2 = 0 + 0.000 000 189 989 804 390 4;
  • 9) 0.000 000 189 989 804 390 4 × 2 = 0 + 0.000 000 379 979 608 780 8;
  • 10) 0.000 000 379 979 608 780 8 × 2 = 0 + 0.000 000 759 959 217 561 6;
  • 11) 0.000 000 759 959 217 561 6 × 2 = 0 + 0.000 001 519 918 435 123 2;
  • 12) 0.000 001 519 918 435 123 2 × 2 = 0 + 0.000 003 039 836 870 246 4;
  • 13) 0.000 003 039 836 870 246 4 × 2 = 0 + 0.000 006 079 673 740 492 8;
  • 14) 0.000 006 079 673 740 492 8 × 2 = 0 + 0.000 012 159 347 480 985 6;
  • 15) 0.000 012 159 347 480 985 6 × 2 = 0 + 0.000 024 318 694 961 971 2;
  • 16) 0.000 024 318 694 961 971 2 × 2 = 0 + 0.000 048 637 389 923 942 4;
  • 17) 0.000 048 637 389 923 942 4 × 2 = 0 + 0.000 097 274 779 847 884 8;
  • 18) 0.000 097 274 779 847 884 8 × 2 = 0 + 0.000 194 549 559 695 769 6;
  • 19) 0.000 194 549 559 695 769 6 × 2 = 0 + 0.000 389 099 119 391 539 2;
  • 20) 0.000 389 099 119 391 539 2 × 2 = 0 + 0.000 778 198 238 783 078 4;
  • 21) 0.000 778 198 238 783 078 4 × 2 = 0 + 0.001 556 396 477 566 156 8;
  • 22) 0.001 556 396 477 566 156 8 × 2 = 0 + 0.003 112 792 955 132 313 6;
  • 23) 0.003 112 792 955 132 313 6 × 2 = 0 + 0.006 225 585 910 264 627 2;
  • 24) 0.006 225 585 910 264 627 2 × 2 = 0 + 0.012 451 171 820 529 254 4;
  • 25) 0.012 451 171 820 529 254 4 × 2 = 0 + 0.024 902 343 641 058 508 8;
  • 26) 0.024 902 343 641 058 508 8 × 2 = 0 + 0.049 804 687 282 117 017 6;
  • 27) 0.049 804 687 282 117 017 6 × 2 = 0 + 0.099 609 374 564 234 035 2;
  • 28) 0.099 609 374 564 234 035 2 × 2 = 0 + 0.199 218 749 128 468 070 4;
  • 29) 0.199 218 749 128 468 070 4 × 2 = 0 + 0.398 437 498 256 936 140 8;
  • 30) 0.398 437 498 256 936 140 8 × 2 = 0 + 0.796 874 996 513 872 281 6;
  • 31) 0.796 874 996 513 872 281 6 × 2 = 1 + 0.593 749 993 027 744 563 2;
  • 32) 0.593 749 993 027 744 563 2 × 2 = 1 + 0.187 499 986 055 489 126 4;
  • 33) 0.187 499 986 055 489 126 4 × 2 = 0 + 0.374 999 972 110 978 252 8;
  • 34) 0.374 999 972 110 978 252 8 × 2 = 0 + 0.749 999 944 221 956 505 6;
  • 35) 0.749 999 944 221 956 505 6 × 2 = 1 + 0.499 999 888 443 913 011 2;
  • 36) 0.499 999 888 443 913 011 2 × 2 = 0 + 0.999 999 776 887 826 022 4;
  • 37) 0.999 999 776 887 826 022 4 × 2 = 1 + 0.999 999 553 775 652 044 8;
  • 38) 0.999 999 553 775 652 044 8 × 2 = 1 + 0.999 999 107 551 304 089 6;
  • 39) 0.999 999 107 551 304 089 6 × 2 = 1 + 0.999 998 215 102 608 179 2;
  • 40) 0.999 998 215 102 608 179 2 × 2 = 1 + 0.999 996 430 205 216 358 4;
  • 41) 0.999 996 430 205 216 358 4 × 2 = 1 + 0.999 992 860 410 432 716 8;
  • 42) 0.999 992 860 410 432 716 8 × 2 = 1 + 0.999 985 720 820 865 433 6;
  • 43) 0.999 985 720 820 865 433 6 × 2 = 1 + 0.999 971 441 641 730 867 2;
  • 44) 0.999 971 441 641 730 867 2 × 2 = 1 + 0.999 942 883 283 461 734 4;
  • 45) 0.999 942 883 283 461 734 4 × 2 = 1 + 0.999 885 766 566 923 468 8;
  • 46) 0.999 885 766 566 923 468 8 × 2 = 1 + 0.999 771 533 133 846 937 6;
  • 47) 0.999 771 533 133 846 937 6 × 2 = 1 + 0.999 543 066 267 693 875 2;
  • 48) 0.999 543 066 267 693 875 2 × 2 = 1 + 0.999 086 132 535 387 750 4;
  • 49) 0.999 086 132 535 387 750 4 × 2 = 1 + 0.998 172 265 070 775 500 8;
  • 50) 0.998 172 265 070 775 500 8 × 2 = 1 + 0.996 344 530 141 551 001 6;
  • 51) 0.996 344 530 141 551 001 6 × 2 = 1 + 0.992 689 060 283 102 003 2;
  • 52) 0.992 689 060 283 102 003 2 × 2 = 1 + 0.985 378 120 566 204 006 4;
  • 53) 0.985 378 120 566 204 006 4 × 2 = 1 + 0.970 756 241 132 408 012 8;
  • 54) 0.970 756 241 132 408 012 8 × 2 = 1 + 0.941 512 482 264 816 025 6;

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:


0.000 000 000 742 147 673 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 673 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(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:


0.000 000 000 742 147 673 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 1111 111


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 1111 1111 =

100 1011 1111 1111 1111 1111


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 1111 1111


Decimal number -0.000 000 000 742 147 673 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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