-0.000 000 000 742 147 669 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 669 1(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
-0.000 000 000 742 147 669 1(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:

|-0.000 000 000 742 147 669 1| = 0.000 000 000 742 147 669 1


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: 0.000 000 000 742 147 669 1.

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) 0.000 000 000 742 147 669 1 × 2 = 0 + 0.000 000 001 484 295 338 2;
  • 2) 0.000 000 001 484 295 338 2 × 2 = 0 + 0.000 000 002 968 590 676 4;
  • 3) 0.000 000 002 968 590 676 4 × 2 = 0 + 0.000 000 005 937 181 352 8;
  • 4) 0.000 000 005 937 181 352 8 × 2 = 0 + 0.000 000 011 874 362 705 6;
  • 5) 0.000 000 011 874 362 705 6 × 2 = 0 + 0.000 000 023 748 725 411 2;
  • 6) 0.000 000 023 748 725 411 2 × 2 = 0 + 0.000 000 047 497 450 822 4;
  • 7) 0.000 000 047 497 450 822 4 × 2 = 0 + 0.000 000 094 994 901 644 8;
  • 8) 0.000 000 094 994 901 644 8 × 2 = 0 + 0.000 000 189 989 803 289 6;
  • 9) 0.000 000 189 989 803 289 6 × 2 = 0 + 0.000 000 379 979 606 579 2;
  • 10) 0.000 000 379 979 606 579 2 × 2 = 0 + 0.000 000 759 959 213 158 4;
  • 11) 0.000 000 759 959 213 158 4 × 2 = 0 + 0.000 001 519 918 426 316 8;
  • 12) 0.000 001 519 918 426 316 8 × 2 = 0 + 0.000 003 039 836 852 633 6;
  • 13) 0.000 003 039 836 852 633 6 × 2 = 0 + 0.000 006 079 673 705 267 2;
  • 14) 0.000 006 079 673 705 267 2 × 2 = 0 + 0.000 012 159 347 410 534 4;
  • 15) 0.000 012 159 347 410 534 4 × 2 = 0 + 0.000 024 318 694 821 068 8;
  • 16) 0.000 024 318 694 821 068 8 × 2 = 0 + 0.000 048 637 389 642 137 6;
  • 17) 0.000 048 637 389 642 137 6 × 2 = 0 + 0.000 097 274 779 284 275 2;
  • 18) 0.000 097 274 779 284 275 2 × 2 = 0 + 0.000 194 549 558 568 550 4;
  • 19) 0.000 194 549 558 568 550 4 × 2 = 0 + 0.000 389 099 117 137 100 8;
  • 20) 0.000 389 099 117 137 100 8 × 2 = 0 + 0.000 778 198 234 274 201 6;
  • 21) 0.000 778 198 234 274 201 6 × 2 = 0 + 0.001 556 396 468 548 403 2;
  • 22) 0.001 556 396 468 548 403 2 × 2 = 0 + 0.003 112 792 937 096 806 4;
  • 23) 0.003 112 792 937 096 806 4 × 2 = 0 + 0.006 225 585 874 193 612 8;
  • 24) 0.006 225 585 874 193 612 8 × 2 = 0 + 0.012 451 171 748 387 225 6;
  • 25) 0.012 451 171 748 387 225 6 × 2 = 0 + 0.024 902 343 496 774 451 2;
  • 26) 0.024 902 343 496 774 451 2 × 2 = 0 + 0.049 804 686 993 548 902 4;
  • 27) 0.049 804 686 993 548 902 4 × 2 = 0 + 0.099 609 373 987 097 804 8;
  • 28) 0.099 609 373 987 097 804 8 × 2 = 0 + 0.199 218 747 974 195 609 6;
  • 29) 0.199 218 747 974 195 609 6 × 2 = 0 + 0.398 437 495 948 391 219 2;
  • 30) 0.398 437 495 948 391 219 2 × 2 = 0 + 0.796 874 991 896 782 438 4;
  • 31) 0.796 874 991 896 782 438 4 × 2 = 1 + 0.593 749 983 793 564 876 8;
  • 32) 0.593 749 983 793 564 876 8 × 2 = 1 + 0.187 499 967 587 129 753 6;
  • 33) 0.187 499 967 587 129 753 6 × 2 = 0 + 0.374 999 935 174 259 507 2;
  • 34) 0.374 999 935 174 259 507 2 × 2 = 0 + 0.749 999 870 348 519 014 4;
  • 35) 0.749 999 870 348 519 014 4 × 2 = 1 + 0.499 999 740 697 038 028 8;
  • 36) 0.499 999 740 697 038 028 8 × 2 = 0 + 0.999 999 481 394 076 057 6;
  • 37) 0.999 999 481 394 076 057 6 × 2 = 1 + 0.999 998 962 788 152 115 2;
  • 38) 0.999 998 962 788 152 115 2 × 2 = 1 + 0.999 997 925 576 304 230 4;
  • 39) 0.999 997 925 576 304 230 4 × 2 = 1 + 0.999 995 851 152 608 460 8;
  • 40) 0.999 995 851 152 608 460 8 × 2 = 1 + 0.999 991 702 305 216 921 6;
  • 41) 0.999 991 702 305 216 921 6 × 2 = 1 + 0.999 983 404 610 433 843 2;
  • 42) 0.999 983 404 610 433 843 2 × 2 = 1 + 0.999 966 809 220 867 686 4;
  • 43) 0.999 966 809 220 867 686 4 × 2 = 1 + 0.999 933 618 441 735 372 8;
  • 44) 0.999 933 618 441 735 372 8 × 2 = 1 + 0.999 867 236 883 470 745 6;
  • 45) 0.999 867 236 883 470 745 6 × 2 = 1 + 0.999 734 473 766 941 491 2;
  • 46) 0.999 734 473 766 941 491 2 × 2 = 1 + 0.999 468 947 533 882 982 4;
  • 47) 0.999 468 947 533 882 982 4 × 2 = 1 + 0.998 937 895 067 765 964 8;
  • 48) 0.998 937 895 067 765 964 8 × 2 = 1 + 0.997 875 790 135 531 929 6;
  • 49) 0.997 875 790 135 531 929 6 × 2 = 1 + 0.995 751 580 271 063 859 2;
  • 50) 0.995 751 580 271 063 859 2 × 2 = 1 + 0.991 503 160 542 127 718 4;
  • 51) 0.991 503 160 542 127 718 4 × 2 = 1 + 0.983 006 321 084 255 436 8;
  • 52) 0.983 006 321 084 255 436 8 × 2 = 1 + 0.966 012 642 168 510 873 6;
  • 53) 0.966 012 642 168 510 873 6 × 2 = 1 + 0.932 025 284 337 021 747 2;
  • 54) 0.932 025 284 337 021 747 2 × 2 = 1 + 0.864 050 568 674 043 494 4;

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:


0.000 000 000 742 147 669 1(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 669 1(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:


0.000 000 000 742 147 669 1(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 -0.000 000 000 742 147 669 1 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