-0.000 000 000 742 147 676 646 614 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 614(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
-0.000 000 000 742 147 676 646 614(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:

|-0.000 000 000 742 147 676 646 614| = 0.000 000 000 742 147 676 646 614


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: 0.000 000 000 742 147 676 646 614.

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) 0.000 000 000 742 147 676 646 614 × 2 = 0 + 0.000 000 001 484 295 353 293 228;
  • 2) 0.000 000 001 484 295 353 293 228 × 2 = 0 + 0.000 000 002 968 590 706 586 456;
  • 3) 0.000 000 002 968 590 706 586 456 × 2 = 0 + 0.000 000 005 937 181 413 172 912;
  • 4) 0.000 000 005 937 181 413 172 912 × 2 = 0 + 0.000 000 011 874 362 826 345 824;
  • 5) 0.000 000 011 874 362 826 345 824 × 2 = 0 + 0.000 000 023 748 725 652 691 648;
  • 6) 0.000 000 023 748 725 652 691 648 × 2 = 0 + 0.000 000 047 497 451 305 383 296;
  • 7) 0.000 000 047 497 451 305 383 296 × 2 = 0 + 0.000 000 094 994 902 610 766 592;
  • 8) 0.000 000 094 994 902 610 766 592 × 2 = 0 + 0.000 000 189 989 805 221 533 184;
  • 9) 0.000 000 189 989 805 221 533 184 × 2 = 0 + 0.000 000 379 979 610 443 066 368;
  • 10) 0.000 000 379 979 610 443 066 368 × 2 = 0 + 0.000 000 759 959 220 886 132 736;
  • 11) 0.000 000 759 959 220 886 132 736 × 2 = 0 + 0.000 001 519 918 441 772 265 472;
  • 12) 0.000 001 519 918 441 772 265 472 × 2 = 0 + 0.000 003 039 836 883 544 530 944;
  • 13) 0.000 003 039 836 883 544 530 944 × 2 = 0 + 0.000 006 079 673 767 089 061 888;
  • 14) 0.000 006 079 673 767 089 061 888 × 2 = 0 + 0.000 012 159 347 534 178 123 776;
  • 15) 0.000 012 159 347 534 178 123 776 × 2 = 0 + 0.000 024 318 695 068 356 247 552;
  • 16) 0.000 024 318 695 068 356 247 552 × 2 = 0 + 0.000 048 637 390 136 712 495 104;
  • 17) 0.000 048 637 390 136 712 495 104 × 2 = 0 + 0.000 097 274 780 273 424 990 208;
  • 18) 0.000 097 274 780 273 424 990 208 × 2 = 0 + 0.000 194 549 560 546 849 980 416;
  • 19) 0.000 194 549 560 546 849 980 416 × 2 = 0 + 0.000 389 099 121 093 699 960 832;
  • 20) 0.000 389 099 121 093 699 960 832 × 2 = 0 + 0.000 778 198 242 187 399 921 664;
  • 21) 0.000 778 198 242 187 399 921 664 × 2 = 0 + 0.001 556 396 484 374 799 843 328;
  • 22) 0.001 556 396 484 374 799 843 328 × 2 = 0 + 0.003 112 792 968 749 599 686 656;
  • 23) 0.003 112 792 968 749 599 686 656 × 2 = 0 + 0.006 225 585 937 499 199 373 312;
  • 24) 0.006 225 585 937 499 199 373 312 × 2 = 0 + 0.012 451 171 874 998 398 746 624;
  • 25) 0.012 451 171 874 998 398 746 624 × 2 = 0 + 0.024 902 343 749 996 797 493 248;
  • 26) 0.024 902 343 749 996 797 493 248 × 2 = 0 + 0.049 804 687 499 993 594 986 496;
  • 27) 0.049 804 687 499 993 594 986 496 × 2 = 0 + 0.099 609 374 999 987 189 972 992;
  • 28) 0.099 609 374 999 987 189 972 992 × 2 = 0 + 0.199 218 749 999 974 379 945 984;
  • 29) 0.199 218 749 999 974 379 945 984 × 2 = 0 + 0.398 437 499 999 948 759 891 968;
  • 30) 0.398 437 499 999 948 759 891 968 × 2 = 0 + 0.796 874 999 999 897 519 783 936;
  • 31) 0.796 874 999 999 897 519 783 936 × 2 = 1 + 0.593 749 999 999 795 039 567 872;
  • 32) 0.593 749 999 999 795 039 567 872 × 2 = 1 + 0.187 499 999 999 590 079 135 744;
  • 33) 0.187 499 999 999 590 079 135 744 × 2 = 0 + 0.374 999 999 999 180 158 271 488;
  • 34) 0.374 999 999 999 180 158 271 488 × 2 = 0 + 0.749 999 999 998 360 316 542 976;
  • 35) 0.749 999 999 998 360 316 542 976 × 2 = 1 + 0.499 999 999 996 720 633 085 952;
  • 36) 0.499 999 999 996 720 633 085 952 × 2 = 0 + 0.999 999 999 993 441 266 171 904;
  • 37) 0.999 999 999 993 441 266 171 904 × 2 = 1 + 0.999 999 999 986 882 532 343 808;
  • 38) 0.999 999 999 986 882 532 343 808 × 2 = 1 + 0.999 999 999 973 765 064 687 616;
  • 39) 0.999 999 999 973 765 064 687 616 × 2 = 1 + 0.999 999 999 947 530 129 375 232;
  • 40) 0.999 999 999 947 530 129 375 232 × 2 = 1 + 0.999 999 999 895 060 258 750 464;
  • 41) 0.999 999 999 895 060 258 750 464 × 2 = 1 + 0.999 999 999 790 120 517 500 928;
  • 42) 0.999 999 999 790 120 517 500 928 × 2 = 1 + 0.999 999 999 580 241 035 001 856;
  • 43) 0.999 999 999 580 241 035 001 856 × 2 = 1 + 0.999 999 999 160 482 070 003 712;
  • 44) 0.999 999 999 160 482 070 003 712 × 2 = 1 + 0.999 999 998 320 964 140 007 424;
  • 45) 0.999 999 998 320 964 140 007 424 × 2 = 1 + 0.999 999 996 641 928 280 014 848;
  • 46) 0.999 999 996 641 928 280 014 848 × 2 = 1 + 0.999 999 993 283 856 560 029 696;
  • 47) 0.999 999 993 283 856 560 029 696 × 2 = 1 + 0.999 999 986 567 713 120 059 392;
  • 48) 0.999 999 986 567 713 120 059 392 × 2 = 1 + 0.999 999 973 135 426 240 118 784;
  • 49) 0.999 999 973 135 426 240 118 784 × 2 = 1 + 0.999 999 946 270 852 480 237 568;
  • 50) 0.999 999 946 270 852 480 237 568 × 2 = 1 + 0.999 999 892 541 704 960 475 136;
  • 51) 0.999 999 892 541 704 960 475 136 × 2 = 1 + 0.999 999 785 083 409 920 950 272;
  • 52) 0.999 999 785 083 409 920 950 272 × 2 = 1 + 0.999 999 570 166 819 841 900 544;
  • 53) 0.999 999 570 166 819 841 900 544 × 2 = 1 + 0.999 999 140 333 639 683 801 088;
  • 54) 0.999 999 140 333 639 683 801 088 × 2 = 1 + 0.999 998 280 667 279 367 602 176;

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:


0.000 000 000 742 147 676 646 614(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 614(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:


0.000 000 000 742 147 676 646 614(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 -0.000 000 000 742 147 676 646 614 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