-0.000 000 000 742 147 677 44 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 677 44(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
-0.000 000 000 742 147 677 44(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:

|-0.000 000 000 742 147 677 44| = 0.000 000 000 742 147 677 44


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: 0.000 000 000 742 147 677 44.

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) 0.000 000 000 742 147 677 44 × 2 = 0 + 0.000 000 001 484 295 354 88;
  • 2) 0.000 000 001 484 295 354 88 × 2 = 0 + 0.000 000 002 968 590 709 76;
  • 3) 0.000 000 002 968 590 709 76 × 2 = 0 + 0.000 000 005 937 181 419 52;
  • 4) 0.000 000 005 937 181 419 52 × 2 = 0 + 0.000 000 011 874 362 839 04;
  • 5) 0.000 000 011 874 362 839 04 × 2 = 0 + 0.000 000 023 748 725 678 08;
  • 6) 0.000 000 023 748 725 678 08 × 2 = 0 + 0.000 000 047 497 451 356 16;
  • 7) 0.000 000 047 497 451 356 16 × 2 = 0 + 0.000 000 094 994 902 712 32;
  • 8) 0.000 000 094 994 902 712 32 × 2 = 0 + 0.000 000 189 989 805 424 64;
  • 9) 0.000 000 189 989 805 424 64 × 2 = 0 + 0.000 000 379 979 610 849 28;
  • 10) 0.000 000 379 979 610 849 28 × 2 = 0 + 0.000 000 759 959 221 698 56;
  • 11) 0.000 000 759 959 221 698 56 × 2 = 0 + 0.000 001 519 918 443 397 12;
  • 12) 0.000 001 519 918 443 397 12 × 2 = 0 + 0.000 003 039 836 886 794 24;
  • 13) 0.000 003 039 836 886 794 24 × 2 = 0 + 0.000 006 079 673 773 588 48;
  • 14) 0.000 006 079 673 773 588 48 × 2 = 0 + 0.000 012 159 347 547 176 96;
  • 15) 0.000 012 159 347 547 176 96 × 2 = 0 + 0.000 024 318 695 094 353 92;
  • 16) 0.000 024 318 695 094 353 92 × 2 = 0 + 0.000 048 637 390 188 707 84;
  • 17) 0.000 048 637 390 188 707 84 × 2 = 0 + 0.000 097 274 780 377 415 68;
  • 18) 0.000 097 274 780 377 415 68 × 2 = 0 + 0.000 194 549 560 754 831 36;
  • 19) 0.000 194 549 560 754 831 36 × 2 = 0 + 0.000 389 099 121 509 662 72;
  • 20) 0.000 389 099 121 509 662 72 × 2 = 0 + 0.000 778 198 243 019 325 44;
  • 21) 0.000 778 198 243 019 325 44 × 2 = 0 + 0.001 556 396 486 038 650 88;
  • 22) 0.001 556 396 486 038 650 88 × 2 = 0 + 0.003 112 792 972 077 301 76;
  • 23) 0.003 112 792 972 077 301 76 × 2 = 0 + 0.006 225 585 944 154 603 52;
  • 24) 0.006 225 585 944 154 603 52 × 2 = 0 + 0.012 451 171 888 309 207 04;
  • 25) 0.012 451 171 888 309 207 04 × 2 = 0 + 0.024 902 343 776 618 414 08;
  • 26) 0.024 902 343 776 618 414 08 × 2 = 0 + 0.049 804 687 553 236 828 16;
  • 27) 0.049 804 687 553 236 828 16 × 2 = 0 + 0.099 609 375 106 473 656 32;
  • 28) 0.099 609 375 106 473 656 32 × 2 = 0 + 0.199 218 750 212 947 312 64;
  • 29) 0.199 218 750 212 947 312 64 × 2 = 0 + 0.398 437 500 425 894 625 28;
  • 30) 0.398 437 500 425 894 625 28 × 2 = 0 + 0.796 875 000 851 789 250 56;
  • 31) 0.796 875 000 851 789 250 56 × 2 = 1 + 0.593 750 001 703 578 501 12;
  • 32) 0.593 750 001 703 578 501 12 × 2 = 1 + 0.187 500 003 407 157 002 24;
  • 33) 0.187 500 003 407 157 002 24 × 2 = 0 + 0.375 000 006 814 314 004 48;
  • 34) 0.375 000 006 814 314 004 48 × 2 = 0 + 0.750 000 013 628 628 008 96;
  • 35) 0.750 000 013 628 628 008 96 × 2 = 1 + 0.500 000 027 257 256 017 92;
  • 36) 0.500 000 027 257 256 017 92 × 2 = 1 + 0.000 000 054 514 512 035 84;
  • 37) 0.000 000 054 514 512 035 84 × 2 = 0 + 0.000 000 109 029 024 071 68;
  • 38) 0.000 000 109 029 024 071 68 × 2 = 0 + 0.000 000 218 058 048 143 36;
  • 39) 0.000 000 218 058 048 143 36 × 2 = 0 + 0.000 000 436 116 096 286 72;
  • 40) 0.000 000 436 116 096 286 72 × 2 = 0 + 0.000 000 872 232 192 573 44;
  • 41) 0.000 000 872 232 192 573 44 × 2 = 0 + 0.000 001 744 464 385 146 88;
  • 42) 0.000 001 744 464 385 146 88 × 2 = 0 + 0.000 003 488 928 770 293 76;
  • 43) 0.000 003 488 928 770 293 76 × 2 = 0 + 0.000 006 977 857 540 587 52;
  • 44) 0.000 006 977 857 540 587 52 × 2 = 0 + 0.000 013 955 715 081 175 04;
  • 45) 0.000 013 955 715 081 175 04 × 2 = 0 + 0.000 027 911 430 162 350 08;
  • 46) 0.000 027 911 430 162 350 08 × 2 = 0 + 0.000 055 822 860 324 700 16;
  • 47) 0.000 055 822 860 324 700 16 × 2 = 0 + 0.000 111 645 720 649 400 32;
  • 48) 0.000 111 645 720 649 400 32 × 2 = 0 + 0.000 223 291 441 298 800 64;
  • 49) 0.000 223 291 441 298 800 64 × 2 = 0 + 0.000 446 582 882 597 601 28;
  • 50) 0.000 446 582 882 597 601 28 × 2 = 0 + 0.000 893 165 765 195 202 56;
  • 51) 0.000 893 165 765 195 202 56 × 2 = 0 + 0.001 786 331 530 390 405 12;
  • 52) 0.001 786 331 530 390 405 12 × 2 = 0 + 0.003 572 663 060 780 810 24;
  • 53) 0.003 572 663 060 780 810 24 × 2 = 0 + 0.007 145 326 121 561 620 48;
  • 54) 0.007 145 326 121 561 620 48 × 2 = 0 + 0.014 290 652 243 123 240 96;

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:


0.000 000 000 742 147 677 44(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 677 44(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:


0.000 000 000 742 147 677 44(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 -0.000 000 000 742 147 677 44 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