-0.000 000 000 742 147 676 646 707 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 707 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
-0.000 000 000 742 147 676 646 707 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:

|-0.000 000 000 742 147 676 646 707 9| = 0.000 000 000 742 147 676 646 707 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: 0.000 000 000 742 147 676 646 707 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) 0.000 000 000 742 147 676 646 707 9 × 2 = 0 + 0.000 000 001 484 295 353 293 415 8;
  • 2) 0.000 000 001 484 295 353 293 415 8 × 2 = 0 + 0.000 000 002 968 590 706 586 831 6;
  • 3) 0.000 000 002 968 590 706 586 831 6 × 2 = 0 + 0.000 000 005 937 181 413 173 663 2;
  • 4) 0.000 000 005 937 181 413 173 663 2 × 2 = 0 + 0.000 000 011 874 362 826 347 326 4;
  • 5) 0.000 000 011 874 362 826 347 326 4 × 2 = 0 + 0.000 000 023 748 725 652 694 652 8;
  • 6) 0.000 000 023 748 725 652 694 652 8 × 2 = 0 + 0.000 000 047 497 451 305 389 305 6;
  • 7) 0.000 000 047 497 451 305 389 305 6 × 2 = 0 + 0.000 000 094 994 902 610 778 611 2;
  • 8) 0.000 000 094 994 902 610 778 611 2 × 2 = 0 + 0.000 000 189 989 805 221 557 222 4;
  • 9) 0.000 000 189 989 805 221 557 222 4 × 2 = 0 + 0.000 000 379 979 610 443 114 444 8;
  • 10) 0.000 000 379 979 610 443 114 444 8 × 2 = 0 + 0.000 000 759 959 220 886 228 889 6;
  • 11) 0.000 000 759 959 220 886 228 889 6 × 2 = 0 + 0.000 001 519 918 441 772 457 779 2;
  • 12) 0.000 001 519 918 441 772 457 779 2 × 2 = 0 + 0.000 003 039 836 883 544 915 558 4;
  • 13) 0.000 003 039 836 883 544 915 558 4 × 2 = 0 + 0.000 006 079 673 767 089 831 116 8;
  • 14) 0.000 006 079 673 767 089 831 116 8 × 2 = 0 + 0.000 012 159 347 534 179 662 233 6;
  • 15) 0.000 012 159 347 534 179 662 233 6 × 2 = 0 + 0.000 024 318 695 068 359 324 467 2;
  • 16) 0.000 024 318 695 068 359 324 467 2 × 2 = 0 + 0.000 048 637 390 136 718 648 934 4;
  • 17) 0.000 048 637 390 136 718 648 934 4 × 2 = 0 + 0.000 097 274 780 273 437 297 868 8;
  • 18) 0.000 097 274 780 273 437 297 868 8 × 2 = 0 + 0.000 194 549 560 546 874 595 737 6;
  • 19) 0.000 194 549 560 546 874 595 737 6 × 2 = 0 + 0.000 389 099 121 093 749 191 475 2;
  • 20) 0.000 389 099 121 093 749 191 475 2 × 2 = 0 + 0.000 778 198 242 187 498 382 950 4;
  • 21) 0.000 778 198 242 187 498 382 950 4 × 2 = 0 + 0.001 556 396 484 374 996 765 900 8;
  • 22) 0.001 556 396 484 374 996 765 900 8 × 2 = 0 + 0.003 112 792 968 749 993 531 801 6;
  • 23) 0.003 112 792 968 749 993 531 801 6 × 2 = 0 + 0.006 225 585 937 499 987 063 603 2;
  • 24) 0.006 225 585 937 499 987 063 603 2 × 2 = 0 + 0.012 451 171 874 999 974 127 206 4;
  • 25) 0.012 451 171 874 999 974 127 206 4 × 2 = 0 + 0.024 902 343 749 999 948 254 412 8;
  • 26) 0.024 902 343 749 999 948 254 412 8 × 2 = 0 + 0.049 804 687 499 999 896 508 825 6;
  • 27) 0.049 804 687 499 999 896 508 825 6 × 2 = 0 + 0.099 609 374 999 999 793 017 651 2;
  • 28) 0.099 609 374 999 999 793 017 651 2 × 2 = 0 + 0.199 218 749 999 999 586 035 302 4;
  • 29) 0.199 218 749 999 999 586 035 302 4 × 2 = 0 + 0.398 437 499 999 999 172 070 604 8;
  • 30) 0.398 437 499 999 999 172 070 604 8 × 2 = 0 + 0.796 874 999 999 998 344 141 209 6;
  • 31) 0.796 874 999 999 998 344 141 209 6 × 2 = 1 + 0.593 749 999 999 996 688 282 419 2;
  • 32) 0.593 749 999 999 996 688 282 419 2 × 2 = 1 + 0.187 499 999 999 993 376 564 838 4;
  • 33) 0.187 499 999 999 993 376 564 838 4 × 2 = 0 + 0.374 999 999 999 986 753 129 676 8;
  • 34) 0.374 999 999 999 986 753 129 676 8 × 2 = 0 + 0.749 999 999 999 973 506 259 353 6;
  • 35) 0.749 999 999 999 973 506 259 353 6 × 2 = 1 + 0.499 999 999 999 947 012 518 707 2;
  • 36) 0.499 999 999 999 947 012 518 707 2 × 2 = 0 + 0.999 999 999 999 894 025 037 414 4;
  • 37) 0.999 999 999 999 894 025 037 414 4 × 2 = 1 + 0.999 999 999 999 788 050 074 828 8;
  • 38) 0.999 999 999 999 788 050 074 828 8 × 2 = 1 + 0.999 999 999 999 576 100 149 657 6;
  • 39) 0.999 999 999 999 576 100 149 657 6 × 2 = 1 + 0.999 999 999 999 152 200 299 315 2;
  • 40) 0.999 999 999 999 152 200 299 315 2 × 2 = 1 + 0.999 999 999 998 304 400 598 630 4;
  • 41) 0.999 999 999 998 304 400 598 630 4 × 2 = 1 + 0.999 999 999 996 608 801 197 260 8;
  • 42) 0.999 999 999 996 608 801 197 260 8 × 2 = 1 + 0.999 999 999 993 217 602 394 521 6;
  • 43) 0.999 999 999 993 217 602 394 521 6 × 2 = 1 + 0.999 999 999 986 435 204 789 043 2;
  • 44) 0.999 999 999 986 435 204 789 043 2 × 2 = 1 + 0.999 999 999 972 870 409 578 086 4;
  • 45) 0.999 999 999 972 870 409 578 086 4 × 2 = 1 + 0.999 999 999 945 740 819 156 172 8;
  • 46) 0.999 999 999 945 740 819 156 172 8 × 2 = 1 + 0.999 999 999 891 481 638 312 345 6;
  • 47) 0.999 999 999 891 481 638 312 345 6 × 2 = 1 + 0.999 999 999 782 963 276 624 691 2;
  • 48) 0.999 999 999 782 963 276 624 691 2 × 2 = 1 + 0.999 999 999 565 926 553 249 382 4;
  • 49) 0.999 999 999 565 926 553 249 382 4 × 2 = 1 + 0.999 999 999 131 853 106 498 764 8;
  • 50) 0.999 999 999 131 853 106 498 764 8 × 2 = 1 + 0.999 999 998 263 706 212 997 529 6;
  • 51) 0.999 999 998 263 706 212 997 529 6 × 2 = 1 + 0.999 999 996 527 412 425 995 059 2;
  • 52) 0.999 999 996 527 412 425 995 059 2 × 2 = 1 + 0.999 999 993 054 824 851 990 118 4;
  • 53) 0.999 999 993 054 824 851 990 118 4 × 2 = 1 + 0.999 999 986 109 649 703 980 236 8;
  • 54) 0.999 999 986 109 649 703 980 236 8 × 2 = 1 + 0.999 999 972 219 299 407 960 473 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:


0.000 000 000 742 147 676 646 707 9(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 707 9(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:


0.000 000 000 742 147 676 646 707 9(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 -0.000 000 000 742 147 676 646 707 9 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