-0.000 000 000 742 147 677 71 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 677 71(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 71(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 71| = 0.000 000 000 742 147 677 71


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 71.

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 71 × 2 = 0 + 0.000 000 001 484 295 355 42;
  • 2) 0.000 000 001 484 295 355 42 × 2 = 0 + 0.000 000 002 968 590 710 84;
  • 3) 0.000 000 002 968 590 710 84 × 2 = 0 + 0.000 000 005 937 181 421 68;
  • 4) 0.000 000 005 937 181 421 68 × 2 = 0 + 0.000 000 011 874 362 843 36;
  • 5) 0.000 000 011 874 362 843 36 × 2 = 0 + 0.000 000 023 748 725 686 72;
  • 6) 0.000 000 023 748 725 686 72 × 2 = 0 + 0.000 000 047 497 451 373 44;
  • 7) 0.000 000 047 497 451 373 44 × 2 = 0 + 0.000 000 094 994 902 746 88;
  • 8) 0.000 000 094 994 902 746 88 × 2 = 0 + 0.000 000 189 989 805 493 76;
  • 9) 0.000 000 189 989 805 493 76 × 2 = 0 + 0.000 000 379 979 610 987 52;
  • 10) 0.000 000 379 979 610 987 52 × 2 = 0 + 0.000 000 759 959 221 975 04;
  • 11) 0.000 000 759 959 221 975 04 × 2 = 0 + 0.000 001 519 918 443 950 08;
  • 12) 0.000 001 519 918 443 950 08 × 2 = 0 + 0.000 003 039 836 887 900 16;
  • 13) 0.000 003 039 836 887 900 16 × 2 = 0 + 0.000 006 079 673 775 800 32;
  • 14) 0.000 006 079 673 775 800 32 × 2 = 0 + 0.000 012 159 347 551 600 64;
  • 15) 0.000 012 159 347 551 600 64 × 2 = 0 + 0.000 024 318 695 103 201 28;
  • 16) 0.000 024 318 695 103 201 28 × 2 = 0 + 0.000 048 637 390 206 402 56;
  • 17) 0.000 048 637 390 206 402 56 × 2 = 0 + 0.000 097 274 780 412 805 12;
  • 18) 0.000 097 274 780 412 805 12 × 2 = 0 + 0.000 194 549 560 825 610 24;
  • 19) 0.000 194 549 560 825 610 24 × 2 = 0 + 0.000 389 099 121 651 220 48;
  • 20) 0.000 389 099 121 651 220 48 × 2 = 0 + 0.000 778 198 243 302 440 96;
  • 21) 0.000 778 198 243 302 440 96 × 2 = 0 + 0.001 556 396 486 604 881 92;
  • 22) 0.001 556 396 486 604 881 92 × 2 = 0 + 0.003 112 792 973 209 763 84;
  • 23) 0.003 112 792 973 209 763 84 × 2 = 0 + 0.006 225 585 946 419 527 68;
  • 24) 0.006 225 585 946 419 527 68 × 2 = 0 + 0.012 451 171 892 839 055 36;
  • 25) 0.012 451 171 892 839 055 36 × 2 = 0 + 0.024 902 343 785 678 110 72;
  • 26) 0.024 902 343 785 678 110 72 × 2 = 0 + 0.049 804 687 571 356 221 44;
  • 27) 0.049 804 687 571 356 221 44 × 2 = 0 + 0.099 609 375 142 712 442 88;
  • 28) 0.099 609 375 142 712 442 88 × 2 = 0 + 0.199 218 750 285 424 885 76;
  • 29) 0.199 218 750 285 424 885 76 × 2 = 0 + 0.398 437 500 570 849 771 52;
  • 30) 0.398 437 500 570 849 771 52 × 2 = 0 + 0.796 875 001 141 699 543 04;
  • 31) 0.796 875 001 141 699 543 04 × 2 = 1 + 0.593 750 002 283 399 086 08;
  • 32) 0.593 750 002 283 399 086 08 × 2 = 1 + 0.187 500 004 566 798 172 16;
  • 33) 0.187 500 004 566 798 172 16 × 2 = 0 + 0.375 000 009 133 596 344 32;
  • 34) 0.375 000 009 133 596 344 32 × 2 = 0 + 0.750 000 018 267 192 688 64;
  • 35) 0.750 000 018 267 192 688 64 × 2 = 1 + 0.500 000 036 534 385 377 28;
  • 36) 0.500 000 036 534 385 377 28 × 2 = 1 + 0.000 000 073 068 770 754 56;
  • 37) 0.000 000 073 068 770 754 56 × 2 = 0 + 0.000 000 146 137 541 509 12;
  • 38) 0.000 000 146 137 541 509 12 × 2 = 0 + 0.000 000 292 275 083 018 24;
  • 39) 0.000 000 292 275 083 018 24 × 2 = 0 + 0.000 000 584 550 166 036 48;
  • 40) 0.000 000 584 550 166 036 48 × 2 = 0 + 0.000 001 169 100 332 072 96;
  • 41) 0.000 001 169 100 332 072 96 × 2 = 0 + 0.000 002 338 200 664 145 92;
  • 42) 0.000 002 338 200 664 145 92 × 2 = 0 + 0.000 004 676 401 328 291 84;
  • 43) 0.000 004 676 401 328 291 84 × 2 = 0 + 0.000 009 352 802 656 583 68;
  • 44) 0.000 009 352 802 656 583 68 × 2 = 0 + 0.000 018 705 605 313 167 36;
  • 45) 0.000 018 705 605 313 167 36 × 2 = 0 + 0.000 037 411 210 626 334 72;
  • 46) 0.000 037 411 210 626 334 72 × 2 = 0 + 0.000 074 822 421 252 669 44;
  • 47) 0.000 074 822 421 252 669 44 × 2 = 0 + 0.000 149 644 842 505 338 88;
  • 48) 0.000 149 644 842 505 338 88 × 2 = 0 + 0.000 299 289 685 010 677 76;
  • 49) 0.000 299 289 685 010 677 76 × 2 = 0 + 0.000 598 579 370 021 355 52;
  • 50) 0.000 598 579 370 021 355 52 × 2 = 0 + 0.001 197 158 740 042 711 04;
  • 51) 0.001 197 158 740 042 711 04 × 2 = 0 + 0.002 394 317 480 085 422 08;
  • 52) 0.002 394 317 480 085 422 08 × 2 = 0 + 0.004 788 634 960 170 844 16;
  • 53) 0.004 788 634 960 170 844 16 × 2 = 0 + 0.009 577 269 920 341 688 32;
  • 54) 0.009 577 269 920 341 688 32 × 2 = 0 + 0.019 154 539 840 683 376 64;

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 71(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 71(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 71(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 71 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