-0.000 000 000 742 147 297 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 297(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
-0.000 000 000 742 147 297(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:

|-0.000 000 000 742 147 297| = 0.000 000 000 742 147 297


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: 0.000 000 000 742 147 297.

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) 0.000 000 000 742 147 297 × 2 = 0 + 0.000 000 001 484 294 594;
  • 2) 0.000 000 001 484 294 594 × 2 = 0 + 0.000 000 002 968 589 188;
  • 3) 0.000 000 002 968 589 188 × 2 = 0 + 0.000 000 005 937 178 376;
  • 4) 0.000 000 005 937 178 376 × 2 = 0 + 0.000 000 011 874 356 752;
  • 5) 0.000 000 011 874 356 752 × 2 = 0 + 0.000 000 023 748 713 504;
  • 6) 0.000 000 023 748 713 504 × 2 = 0 + 0.000 000 047 497 427 008;
  • 7) 0.000 000 047 497 427 008 × 2 = 0 + 0.000 000 094 994 854 016;
  • 8) 0.000 000 094 994 854 016 × 2 = 0 + 0.000 000 189 989 708 032;
  • 9) 0.000 000 189 989 708 032 × 2 = 0 + 0.000 000 379 979 416 064;
  • 10) 0.000 000 379 979 416 064 × 2 = 0 + 0.000 000 759 958 832 128;
  • 11) 0.000 000 759 958 832 128 × 2 = 0 + 0.000 001 519 917 664 256;
  • 12) 0.000 001 519 917 664 256 × 2 = 0 + 0.000 003 039 835 328 512;
  • 13) 0.000 003 039 835 328 512 × 2 = 0 + 0.000 006 079 670 657 024;
  • 14) 0.000 006 079 670 657 024 × 2 = 0 + 0.000 012 159 341 314 048;
  • 15) 0.000 012 159 341 314 048 × 2 = 0 + 0.000 024 318 682 628 096;
  • 16) 0.000 024 318 682 628 096 × 2 = 0 + 0.000 048 637 365 256 192;
  • 17) 0.000 048 637 365 256 192 × 2 = 0 + 0.000 097 274 730 512 384;
  • 18) 0.000 097 274 730 512 384 × 2 = 0 + 0.000 194 549 461 024 768;
  • 19) 0.000 194 549 461 024 768 × 2 = 0 + 0.000 389 098 922 049 536;
  • 20) 0.000 389 098 922 049 536 × 2 = 0 + 0.000 778 197 844 099 072;
  • 21) 0.000 778 197 844 099 072 × 2 = 0 + 0.001 556 395 688 198 144;
  • 22) 0.001 556 395 688 198 144 × 2 = 0 + 0.003 112 791 376 396 288;
  • 23) 0.003 112 791 376 396 288 × 2 = 0 + 0.006 225 582 752 792 576;
  • 24) 0.006 225 582 752 792 576 × 2 = 0 + 0.012 451 165 505 585 152;
  • 25) 0.012 451 165 505 585 152 × 2 = 0 + 0.024 902 331 011 170 304;
  • 26) 0.024 902 331 011 170 304 × 2 = 0 + 0.049 804 662 022 340 608;
  • 27) 0.049 804 662 022 340 608 × 2 = 0 + 0.099 609 324 044 681 216;
  • 28) 0.099 609 324 044 681 216 × 2 = 0 + 0.199 218 648 089 362 432;
  • 29) 0.199 218 648 089 362 432 × 2 = 0 + 0.398 437 296 178 724 864;
  • 30) 0.398 437 296 178 724 864 × 2 = 0 + 0.796 874 592 357 449 728;
  • 31) 0.796 874 592 357 449 728 × 2 = 1 + 0.593 749 184 714 899 456;
  • 32) 0.593 749 184 714 899 456 × 2 = 1 + 0.187 498 369 429 798 912;
  • 33) 0.187 498 369 429 798 912 × 2 = 0 + 0.374 996 738 859 597 824;
  • 34) 0.374 996 738 859 597 824 × 2 = 0 + 0.749 993 477 719 195 648;
  • 35) 0.749 993 477 719 195 648 × 2 = 1 + 0.499 986 955 438 391 296;
  • 36) 0.499 986 955 438 391 296 × 2 = 0 + 0.999 973 910 876 782 592;
  • 37) 0.999 973 910 876 782 592 × 2 = 1 + 0.999 947 821 753 565 184;
  • 38) 0.999 947 821 753 565 184 × 2 = 1 + 0.999 895 643 507 130 368;
  • 39) 0.999 895 643 507 130 368 × 2 = 1 + 0.999 791 287 014 260 736;
  • 40) 0.999 791 287 014 260 736 × 2 = 1 + 0.999 582 574 028 521 472;
  • 41) 0.999 582 574 028 521 472 × 2 = 1 + 0.999 165 148 057 042 944;
  • 42) 0.999 165 148 057 042 944 × 2 = 1 + 0.998 330 296 114 085 888;
  • 43) 0.998 330 296 114 085 888 × 2 = 1 + 0.996 660 592 228 171 776;
  • 44) 0.996 660 592 228 171 776 × 2 = 1 + 0.993 321 184 456 343 552;
  • 45) 0.993 321 184 456 343 552 × 2 = 1 + 0.986 642 368 912 687 104;
  • 46) 0.986 642 368 912 687 104 × 2 = 1 + 0.973 284 737 825 374 208;
  • 47) 0.973 284 737 825 374 208 × 2 = 1 + 0.946 569 475 650 748 416;
  • 48) 0.946 569 475 650 748 416 × 2 = 1 + 0.893 138 951 301 496 832;
  • 49) 0.893 138 951 301 496 832 × 2 = 1 + 0.786 277 902 602 993 664;
  • 50) 0.786 277 902 602 993 664 × 2 = 1 + 0.572 555 805 205 987 328;
  • 51) 0.572 555 805 205 987 328 × 2 = 1 + 0.145 111 610 411 974 656;
  • 52) 0.145 111 610 411 974 656 × 2 = 0 + 0.290 223 220 823 949 312;
  • 53) 0.290 223 220 823 949 312 × 2 = 0 + 0.580 446 441 647 898 624;
  • 54) 0.580 446 441 647 898 624 × 2 = 1 + 0.160 892 883 295 797 248;

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:


0.000 000 000 742 147 297(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 01(2)

6. Positive number before normalization:

0.000 000 000 742 147 297(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 01(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:


0.000 000 000 742 147 297(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 01(2) × 20 =

1.1001 0111 1111 1111 1111 001(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 001


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 1001 =

100 1011 1111 1111 1111 1001


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 1001


Decimal number -0.000 000 000 742 147 297 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1001

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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