-0.000 000 000 742 147 676 258 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 258(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
-0.000 000 000 742 147 676 258(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:

|-0.000 000 000 742 147 676 258| = 0.000 000 000 742 147 676 258


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: 0.000 000 000 742 147 676 258.

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) 0.000 000 000 742 147 676 258 × 2 = 0 + 0.000 000 001 484 295 352 516;
  • 2) 0.000 000 001 484 295 352 516 × 2 = 0 + 0.000 000 002 968 590 705 032;
  • 3) 0.000 000 002 968 590 705 032 × 2 = 0 + 0.000 000 005 937 181 410 064;
  • 4) 0.000 000 005 937 181 410 064 × 2 = 0 + 0.000 000 011 874 362 820 128;
  • 5) 0.000 000 011 874 362 820 128 × 2 = 0 + 0.000 000 023 748 725 640 256;
  • 6) 0.000 000 023 748 725 640 256 × 2 = 0 + 0.000 000 047 497 451 280 512;
  • 7) 0.000 000 047 497 451 280 512 × 2 = 0 + 0.000 000 094 994 902 561 024;
  • 8) 0.000 000 094 994 902 561 024 × 2 = 0 + 0.000 000 189 989 805 122 048;
  • 9) 0.000 000 189 989 805 122 048 × 2 = 0 + 0.000 000 379 979 610 244 096;
  • 10) 0.000 000 379 979 610 244 096 × 2 = 0 + 0.000 000 759 959 220 488 192;
  • 11) 0.000 000 759 959 220 488 192 × 2 = 0 + 0.000 001 519 918 440 976 384;
  • 12) 0.000 001 519 918 440 976 384 × 2 = 0 + 0.000 003 039 836 881 952 768;
  • 13) 0.000 003 039 836 881 952 768 × 2 = 0 + 0.000 006 079 673 763 905 536;
  • 14) 0.000 006 079 673 763 905 536 × 2 = 0 + 0.000 012 159 347 527 811 072;
  • 15) 0.000 012 159 347 527 811 072 × 2 = 0 + 0.000 024 318 695 055 622 144;
  • 16) 0.000 024 318 695 055 622 144 × 2 = 0 + 0.000 048 637 390 111 244 288;
  • 17) 0.000 048 637 390 111 244 288 × 2 = 0 + 0.000 097 274 780 222 488 576;
  • 18) 0.000 097 274 780 222 488 576 × 2 = 0 + 0.000 194 549 560 444 977 152;
  • 19) 0.000 194 549 560 444 977 152 × 2 = 0 + 0.000 389 099 120 889 954 304;
  • 20) 0.000 389 099 120 889 954 304 × 2 = 0 + 0.000 778 198 241 779 908 608;
  • 21) 0.000 778 198 241 779 908 608 × 2 = 0 + 0.001 556 396 483 559 817 216;
  • 22) 0.001 556 396 483 559 817 216 × 2 = 0 + 0.003 112 792 967 119 634 432;
  • 23) 0.003 112 792 967 119 634 432 × 2 = 0 + 0.006 225 585 934 239 268 864;
  • 24) 0.006 225 585 934 239 268 864 × 2 = 0 + 0.012 451 171 868 478 537 728;
  • 25) 0.012 451 171 868 478 537 728 × 2 = 0 + 0.024 902 343 736 957 075 456;
  • 26) 0.024 902 343 736 957 075 456 × 2 = 0 + 0.049 804 687 473 914 150 912;
  • 27) 0.049 804 687 473 914 150 912 × 2 = 0 + 0.099 609 374 947 828 301 824;
  • 28) 0.099 609 374 947 828 301 824 × 2 = 0 + 0.199 218 749 895 656 603 648;
  • 29) 0.199 218 749 895 656 603 648 × 2 = 0 + 0.398 437 499 791 313 207 296;
  • 30) 0.398 437 499 791 313 207 296 × 2 = 0 + 0.796 874 999 582 626 414 592;
  • 31) 0.796 874 999 582 626 414 592 × 2 = 1 + 0.593 749 999 165 252 829 184;
  • 32) 0.593 749 999 165 252 829 184 × 2 = 1 + 0.187 499 998 330 505 658 368;
  • 33) 0.187 499 998 330 505 658 368 × 2 = 0 + 0.374 999 996 661 011 316 736;
  • 34) 0.374 999 996 661 011 316 736 × 2 = 0 + 0.749 999 993 322 022 633 472;
  • 35) 0.749 999 993 322 022 633 472 × 2 = 1 + 0.499 999 986 644 045 266 944;
  • 36) 0.499 999 986 644 045 266 944 × 2 = 0 + 0.999 999 973 288 090 533 888;
  • 37) 0.999 999 973 288 090 533 888 × 2 = 1 + 0.999 999 946 576 181 067 776;
  • 38) 0.999 999 946 576 181 067 776 × 2 = 1 + 0.999 999 893 152 362 135 552;
  • 39) 0.999 999 893 152 362 135 552 × 2 = 1 + 0.999 999 786 304 724 271 104;
  • 40) 0.999 999 786 304 724 271 104 × 2 = 1 + 0.999 999 572 609 448 542 208;
  • 41) 0.999 999 572 609 448 542 208 × 2 = 1 + 0.999 999 145 218 897 084 416;
  • 42) 0.999 999 145 218 897 084 416 × 2 = 1 + 0.999 998 290 437 794 168 832;
  • 43) 0.999 998 290 437 794 168 832 × 2 = 1 + 0.999 996 580 875 588 337 664;
  • 44) 0.999 996 580 875 588 337 664 × 2 = 1 + 0.999 993 161 751 176 675 328;
  • 45) 0.999 993 161 751 176 675 328 × 2 = 1 + 0.999 986 323 502 353 350 656;
  • 46) 0.999 986 323 502 353 350 656 × 2 = 1 + 0.999 972 647 004 706 701 312;
  • 47) 0.999 972 647 004 706 701 312 × 2 = 1 + 0.999 945 294 009 413 402 624;
  • 48) 0.999 945 294 009 413 402 624 × 2 = 1 + 0.999 890 588 018 826 805 248;
  • 49) 0.999 890 588 018 826 805 248 × 2 = 1 + 0.999 781 176 037 653 610 496;
  • 50) 0.999 781 176 037 653 610 496 × 2 = 1 + 0.999 562 352 075 307 220 992;
  • 51) 0.999 562 352 075 307 220 992 × 2 = 1 + 0.999 124 704 150 614 441 984;
  • 52) 0.999 124 704 150 614 441 984 × 2 = 1 + 0.998 249 408 301 228 883 968;
  • 53) 0.998 249 408 301 228 883 968 × 2 = 1 + 0.996 498 816 602 457 767 936;
  • 54) 0.996 498 816 602 457 767 936 × 2 = 1 + 0.992 997 633 204 915 535 872;

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:


0.000 000 000 742 147 676 258(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 258(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:


0.000 000 000 742 147 676 258(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 -0.000 000 000 742 147 676 258 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 227 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 minutes

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