-0.000 000 000 742 147 676 648 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 648 6(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
-0.000 000 000 742 147 676 648 6(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:

|-0.000 000 000 742 147 676 648 6| = 0.000 000 000 742 147 676 648 6


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: 0.000 000 000 742 147 676 648 6.

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) 0.000 000 000 742 147 676 648 6 × 2 = 0 + 0.000 000 001 484 295 353 297 2;
  • 2) 0.000 000 001 484 295 353 297 2 × 2 = 0 + 0.000 000 002 968 590 706 594 4;
  • 3) 0.000 000 002 968 590 706 594 4 × 2 = 0 + 0.000 000 005 937 181 413 188 8;
  • 4) 0.000 000 005 937 181 413 188 8 × 2 = 0 + 0.000 000 011 874 362 826 377 6;
  • 5) 0.000 000 011 874 362 826 377 6 × 2 = 0 + 0.000 000 023 748 725 652 755 2;
  • 6) 0.000 000 023 748 725 652 755 2 × 2 = 0 + 0.000 000 047 497 451 305 510 4;
  • 7) 0.000 000 047 497 451 305 510 4 × 2 = 0 + 0.000 000 094 994 902 611 020 8;
  • 8) 0.000 000 094 994 902 611 020 8 × 2 = 0 + 0.000 000 189 989 805 222 041 6;
  • 9) 0.000 000 189 989 805 222 041 6 × 2 = 0 + 0.000 000 379 979 610 444 083 2;
  • 10) 0.000 000 379 979 610 444 083 2 × 2 = 0 + 0.000 000 759 959 220 888 166 4;
  • 11) 0.000 000 759 959 220 888 166 4 × 2 = 0 + 0.000 001 519 918 441 776 332 8;
  • 12) 0.000 001 519 918 441 776 332 8 × 2 = 0 + 0.000 003 039 836 883 552 665 6;
  • 13) 0.000 003 039 836 883 552 665 6 × 2 = 0 + 0.000 006 079 673 767 105 331 2;
  • 14) 0.000 006 079 673 767 105 331 2 × 2 = 0 + 0.000 012 159 347 534 210 662 4;
  • 15) 0.000 012 159 347 534 210 662 4 × 2 = 0 + 0.000 024 318 695 068 421 324 8;
  • 16) 0.000 024 318 695 068 421 324 8 × 2 = 0 + 0.000 048 637 390 136 842 649 6;
  • 17) 0.000 048 637 390 136 842 649 6 × 2 = 0 + 0.000 097 274 780 273 685 299 2;
  • 18) 0.000 097 274 780 273 685 299 2 × 2 = 0 + 0.000 194 549 560 547 370 598 4;
  • 19) 0.000 194 549 560 547 370 598 4 × 2 = 0 + 0.000 389 099 121 094 741 196 8;
  • 20) 0.000 389 099 121 094 741 196 8 × 2 = 0 + 0.000 778 198 242 189 482 393 6;
  • 21) 0.000 778 198 242 189 482 393 6 × 2 = 0 + 0.001 556 396 484 378 964 787 2;
  • 22) 0.001 556 396 484 378 964 787 2 × 2 = 0 + 0.003 112 792 968 757 929 574 4;
  • 23) 0.003 112 792 968 757 929 574 4 × 2 = 0 + 0.006 225 585 937 515 859 148 8;
  • 24) 0.006 225 585 937 515 859 148 8 × 2 = 0 + 0.012 451 171 875 031 718 297 6;
  • 25) 0.012 451 171 875 031 718 297 6 × 2 = 0 + 0.024 902 343 750 063 436 595 2;
  • 26) 0.024 902 343 750 063 436 595 2 × 2 = 0 + 0.049 804 687 500 126 873 190 4;
  • 27) 0.049 804 687 500 126 873 190 4 × 2 = 0 + 0.099 609 375 000 253 746 380 8;
  • 28) 0.099 609 375 000 253 746 380 8 × 2 = 0 + 0.199 218 750 000 507 492 761 6;
  • 29) 0.199 218 750 000 507 492 761 6 × 2 = 0 + 0.398 437 500 001 014 985 523 2;
  • 30) 0.398 437 500 001 014 985 523 2 × 2 = 0 + 0.796 875 000 002 029 971 046 4;
  • 31) 0.796 875 000 002 029 971 046 4 × 2 = 1 + 0.593 750 000 004 059 942 092 8;
  • 32) 0.593 750 000 004 059 942 092 8 × 2 = 1 + 0.187 500 000 008 119 884 185 6;
  • 33) 0.187 500 000 008 119 884 185 6 × 2 = 0 + 0.375 000 000 016 239 768 371 2;
  • 34) 0.375 000 000 016 239 768 371 2 × 2 = 0 + 0.750 000 000 032 479 536 742 4;
  • 35) 0.750 000 000 032 479 536 742 4 × 2 = 1 + 0.500 000 000 064 959 073 484 8;
  • 36) 0.500 000 000 064 959 073 484 8 × 2 = 1 + 0.000 000 000 129 918 146 969 6;
  • 37) 0.000 000 000 129 918 146 969 6 × 2 = 0 + 0.000 000 000 259 836 293 939 2;
  • 38) 0.000 000 000 259 836 293 939 2 × 2 = 0 + 0.000 000 000 519 672 587 878 4;
  • 39) 0.000 000 000 519 672 587 878 4 × 2 = 0 + 0.000 000 001 039 345 175 756 8;
  • 40) 0.000 000 001 039 345 175 756 8 × 2 = 0 + 0.000 000 002 078 690 351 513 6;
  • 41) 0.000 000 002 078 690 351 513 6 × 2 = 0 + 0.000 000 004 157 380 703 027 2;
  • 42) 0.000 000 004 157 380 703 027 2 × 2 = 0 + 0.000 000 008 314 761 406 054 4;
  • 43) 0.000 000 008 314 761 406 054 4 × 2 = 0 + 0.000 000 016 629 522 812 108 8;
  • 44) 0.000 000 016 629 522 812 108 8 × 2 = 0 + 0.000 000 033 259 045 624 217 6;
  • 45) 0.000 000 033 259 045 624 217 6 × 2 = 0 + 0.000 000 066 518 091 248 435 2;
  • 46) 0.000 000 066 518 091 248 435 2 × 2 = 0 + 0.000 000 133 036 182 496 870 4;
  • 47) 0.000 000 133 036 182 496 870 4 × 2 = 0 + 0.000 000 266 072 364 993 740 8;
  • 48) 0.000 000 266 072 364 993 740 8 × 2 = 0 + 0.000 000 532 144 729 987 481 6;
  • 49) 0.000 000 532 144 729 987 481 6 × 2 = 0 + 0.000 001 064 289 459 974 963 2;
  • 50) 0.000 001 064 289 459 974 963 2 × 2 = 0 + 0.000 002 128 578 919 949 926 4;
  • 51) 0.000 002 128 578 919 949 926 4 × 2 = 0 + 0.000 004 257 157 839 899 852 8;
  • 52) 0.000 004 257 157 839 899 852 8 × 2 = 0 + 0.000 008 514 315 679 799 705 6;
  • 53) 0.000 008 514 315 679 799 705 6 × 2 = 0 + 0.000 017 028 631 359 599 411 2;
  • 54) 0.000 017 028 631 359 599 411 2 × 2 = 0 + 0.000 034 057 262 719 198 822 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:


0.000 000 000 742 147 676 648 6(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 648 6(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:


0.000 000 000 742 147 676 648 6(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 -0.000 000 000 742 147 676 648 6 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