-0.000 000 000 742 147 676 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 9(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
-0.000 000 000 742 147 676 9(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:

|-0.000 000 000 742 147 676 9| = 0.000 000 000 742 147 676 9


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: 0.000 000 000 742 147 676 9.

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) 0.000 000 000 742 147 676 9 × 2 = 0 + 0.000 000 001 484 295 353 8;
  • 2) 0.000 000 001 484 295 353 8 × 2 = 0 + 0.000 000 002 968 590 707 6;
  • 3) 0.000 000 002 968 590 707 6 × 2 = 0 + 0.000 000 005 937 181 415 2;
  • 4) 0.000 000 005 937 181 415 2 × 2 = 0 + 0.000 000 011 874 362 830 4;
  • 5) 0.000 000 011 874 362 830 4 × 2 = 0 + 0.000 000 023 748 725 660 8;
  • 6) 0.000 000 023 748 725 660 8 × 2 = 0 + 0.000 000 047 497 451 321 6;
  • 7) 0.000 000 047 497 451 321 6 × 2 = 0 + 0.000 000 094 994 902 643 2;
  • 8) 0.000 000 094 994 902 643 2 × 2 = 0 + 0.000 000 189 989 805 286 4;
  • 9) 0.000 000 189 989 805 286 4 × 2 = 0 + 0.000 000 379 979 610 572 8;
  • 10) 0.000 000 379 979 610 572 8 × 2 = 0 + 0.000 000 759 959 221 145 6;
  • 11) 0.000 000 759 959 221 145 6 × 2 = 0 + 0.000 001 519 918 442 291 2;
  • 12) 0.000 001 519 918 442 291 2 × 2 = 0 + 0.000 003 039 836 884 582 4;
  • 13) 0.000 003 039 836 884 582 4 × 2 = 0 + 0.000 006 079 673 769 164 8;
  • 14) 0.000 006 079 673 769 164 8 × 2 = 0 + 0.000 012 159 347 538 329 6;
  • 15) 0.000 012 159 347 538 329 6 × 2 = 0 + 0.000 024 318 695 076 659 2;
  • 16) 0.000 024 318 695 076 659 2 × 2 = 0 + 0.000 048 637 390 153 318 4;
  • 17) 0.000 048 637 390 153 318 4 × 2 = 0 + 0.000 097 274 780 306 636 8;
  • 18) 0.000 097 274 780 306 636 8 × 2 = 0 + 0.000 194 549 560 613 273 6;
  • 19) 0.000 194 549 560 613 273 6 × 2 = 0 + 0.000 389 099 121 226 547 2;
  • 20) 0.000 389 099 121 226 547 2 × 2 = 0 + 0.000 778 198 242 453 094 4;
  • 21) 0.000 778 198 242 453 094 4 × 2 = 0 + 0.001 556 396 484 906 188 8;
  • 22) 0.001 556 396 484 906 188 8 × 2 = 0 + 0.003 112 792 969 812 377 6;
  • 23) 0.003 112 792 969 812 377 6 × 2 = 0 + 0.006 225 585 939 624 755 2;
  • 24) 0.006 225 585 939 624 755 2 × 2 = 0 + 0.012 451 171 879 249 510 4;
  • 25) 0.012 451 171 879 249 510 4 × 2 = 0 + 0.024 902 343 758 499 020 8;
  • 26) 0.024 902 343 758 499 020 8 × 2 = 0 + 0.049 804 687 516 998 041 6;
  • 27) 0.049 804 687 516 998 041 6 × 2 = 0 + 0.099 609 375 033 996 083 2;
  • 28) 0.099 609 375 033 996 083 2 × 2 = 0 + 0.199 218 750 067 992 166 4;
  • 29) 0.199 218 750 067 992 166 4 × 2 = 0 + 0.398 437 500 135 984 332 8;
  • 30) 0.398 437 500 135 984 332 8 × 2 = 0 + 0.796 875 000 271 968 665 6;
  • 31) 0.796 875 000 271 968 665 6 × 2 = 1 + 0.593 750 000 543 937 331 2;
  • 32) 0.593 750 000 543 937 331 2 × 2 = 1 + 0.187 500 001 087 874 662 4;
  • 33) 0.187 500 001 087 874 662 4 × 2 = 0 + 0.375 000 002 175 749 324 8;
  • 34) 0.375 000 002 175 749 324 8 × 2 = 0 + 0.750 000 004 351 498 649 6;
  • 35) 0.750 000 004 351 498 649 6 × 2 = 1 + 0.500 000 008 702 997 299 2;
  • 36) 0.500 000 008 702 997 299 2 × 2 = 1 + 0.000 000 017 405 994 598 4;
  • 37) 0.000 000 017 405 994 598 4 × 2 = 0 + 0.000 000 034 811 989 196 8;
  • 38) 0.000 000 034 811 989 196 8 × 2 = 0 + 0.000 000 069 623 978 393 6;
  • 39) 0.000 000 069 623 978 393 6 × 2 = 0 + 0.000 000 139 247 956 787 2;
  • 40) 0.000 000 139 247 956 787 2 × 2 = 0 + 0.000 000 278 495 913 574 4;
  • 41) 0.000 000 278 495 913 574 4 × 2 = 0 + 0.000 000 556 991 827 148 8;
  • 42) 0.000 000 556 991 827 148 8 × 2 = 0 + 0.000 001 113 983 654 297 6;
  • 43) 0.000 001 113 983 654 297 6 × 2 = 0 + 0.000 002 227 967 308 595 2;
  • 44) 0.000 002 227 967 308 595 2 × 2 = 0 + 0.000 004 455 934 617 190 4;
  • 45) 0.000 004 455 934 617 190 4 × 2 = 0 + 0.000 008 911 869 234 380 8;
  • 46) 0.000 008 911 869 234 380 8 × 2 = 0 + 0.000 017 823 738 468 761 6;
  • 47) 0.000 017 823 738 468 761 6 × 2 = 0 + 0.000 035 647 476 937 523 2;
  • 48) 0.000 035 647 476 937 523 2 × 2 = 0 + 0.000 071 294 953 875 046 4;
  • 49) 0.000 071 294 953 875 046 4 × 2 = 0 + 0.000 142 589 907 750 092 8;
  • 50) 0.000 142 589 907 750 092 8 × 2 = 0 + 0.000 285 179 815 500 185 6;
  • 51) 0.000 285 179 815 500 185 6 × 2 = 0 + 0.000 570 359 631 000 371 2;
  • 52) 0.000 570 359 631 000 371 2 × 2 = 0 + 0.001 140 719 262 000 742 4;
  • 53) 0.001 140 719 262 000 742 4 × 2 = 0 + 0.002 281 438 524 001 484 8;
  • 54) 0.002 281 438 524 001 484 8 × 2 = 0 + 0.004 562 877 048 002 969 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:


0.000 000 000 742 147 676 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 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:


0.000 000 000 742 147 676 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 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 1000 0000 0000 0000 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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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