-0.000 000 000 742 147 676 646 634 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 634(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
-0.000 000 000 742 147 676 646 634(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:

|-0.000 000 000 742 147 676 646 634| = 0.000 000 000 742 147 676 646 634


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: 0.000 000 000 742 147 676 646 634.

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) 0.000 000 000 742 147 676 646 634 × 2 = 0 + 0.000 000 001 484 295 353 293 268;
  • 2) 0.000 000 001 484 295 353 293 268 × 2 = 0 + 0.000 000 002 968 590 706 586 536;
  • 3) 0.000 000 002 968 590 706 586 536 × 2 = 0 + 0.000 000 005 937 181 413 173 072;
  • 4) 0.000 000 005 937 181 413 173 072 × 2 = 0 + 0.000 000 011 874 362 826 346 144;
  • 5) 0.000 000 011 874 362 826 346 144 × 2 = 0 + 0.000 000 023 748 725 652 692 288;
  • 6) 0.000 000 023 748 725 652 692 288 × 2 = 0 + 0.000 000 047 497 451 305 384 576;
  • 7) 0.000 000 047 497 451 305 384 576 × 2 = 0 + 0.000 000 094 994 902 610 769 152;
  • 8) 0.000 000 094 994 902 610 769 152 × 2 = 0 + 0.000 000 189 989 805 221 538 304;
  • 9) 0.000 000 189 989 805 221 538 304 × 2 = 0 + 0.000 000 379 979 610 443 076 608;
  • 10) 0.000 000 379 979 610 443 076 608 × 2 = 0 + 0.000 000 759 959 220 886 153 216;
  • 11) 0.000 000 759 959 220 886 153 216 × 2 = 0 + 0.000 001 519 918 441 772 306 432;
  • 12) 0.000 001 519 918 441 772 306 432 × 2 = 0 + 0.000 003 039 836 883 544 612 864;
  • 13) 0.000 003 039 836 883 544 612 864 × 2 = 0 + 0.000 006 079 673 767 089 225 728;
  • 14) 0.000 006 079 673 767 089 225 728 × 2 = 0 + 0.000 012 159 347 534 178 451 456;
  • 15) 0.000 012 159 347 534 178 451 456 × 2 = 0 + 0.000 024 318 695 068 356 902 912;
  • 16) 0.000 024 318 695 068 356 902 912 × 2 = 0 + 0.000 048 637 390 136 713 805 824;
  • 17) 0.000 048 637 390 136 713 805 824 × 2 = 0 + 0.000 097 274 780 273 427 611 648;
  • 18) 0.000 097 274 780 273 427 611 648 × 2 = 0 + 0.000 194 549 560 546 855 223 296;
  • 19) 0.000 194 549 560 546 855 223 296 × 2 = 0 + 0.000 389 099 121 093 710 446 592;
  • 20) 0.000 389 099 121 093 710 446 592 × 2 = 0 + 0.000 778 198 242 187 420 893 184;
  • 21) 0.000 778 198 242 187 420 893 184 × 2 = 0 + 0.001 556 396 484 374 841 786 368;
  • 22) 0.001 556 396 484 374 841 786 368 × 2 = 0 + 0.003 112 792 968 749 683 572 736;
  • 23) 0.003 112 792 968 749 683 572 736 × 2 = 0 + 0.006 225 585 937 499 367 145 472;
  • 24) 0.006 225 585 937 499 367 145 472 × 2 = 0 + 0.012 451 171 874 998 734 290 944;
  • 25) 0.012 451 171 874 998 734 290 944 × 2 = 0 + 0.024 902 343 749 997 468 581 888;
  • 26) 0.024 902 343 749 997 468 581 888 × 2 = 0 + 0.049 804 687 499 994 937 163 776;
  • 27) 0.049 804 687 499 994 937 163 776 × 2 = 0 + 0.099 609 374 999 989 874 327 552;
  • 28) 0.099 609 374 999 989 874 327 552 × 2 = 0 + 0.199 218 749 999 979 748 655 104;
  • 29) 0.199 218 749 999 979 748 655 104 × 2 = 0 + 0.398 437 499 999 959 497 310 208;
  • 30) 0.398 437 499 999 959 497 310 208 × 2 = 0 + 0.796 874 999 999 918 994 620 416;
  • 31) 0.796 874 999 999 918 994 620 416 × 2 = 1 + 0.593 749 999 999 837 989 240 832;
  • 32) 0.593 749 999 999 837 989 240 832 × 2 = 1 + 0.187 499 999 999 675 978 481 664;
  • 33) 0.187 499 999 999 675 978 481 664 × 2 = 0 + 0.374 999 999 999 351 956 963 328;
  • 34) 0.374 999 999 999 351 956 963 328 × 2 = 0 + 0.749 999 999 998 703 913 926 656;
  • 35) 0.749 999 999 998 703 913 926 656 × 2 = 1 + 0.499 999 999 997 407 827 853 312;
  • 36) 0.499 999 999 997 407 827 853 312 × 2 = 0 + 0.999 999 999 994 815 655 706 624;
  • 37) 0.999 999 999 994 815 655 706 624 × 2 = 1 + 0.999 999 999 989 631 311 413 248;
  • 38) 0.999 999 999 989 631 311 413 248 × 2 = 1 + 0.999 999 999 979 262 622 826 496;
  • 39) 0.999 999 999 979 262 622 826 496 × 2 = 1 + 0.999 999 999 958 525 245 652 992;
  • 40) 0.999 999 999 958 525 245 652 992 × 2 = 1 + 0.999 999 999 917 050 491 305 984;
  • 41) 0.999 999 999 917 050 491 305 984 × 2 = 1 + 0.999 999 999 834 100 982 611 968;
  • 42) 0.999 999 999 834 100 982 611 968 × 2 = 1 + 0.999 999 999 668 201 965 223 936;
  • 43) 0.999 999 999 668 201 965 223 936 × 2 = 1 + 0.999 999 999 336 403 930 447 872;
  • 44) 0.999 999 999 336 403 930 447 872 × 2 = 1 + 0.999 999 998 672 807 860 895 744;
  • 45) 0.999 999 998 672 807 860 895 744 × 2 = 1 + 0.999 999 997 345 615 721 791 488;
  • 46) 0.999 999 997 345 615 721 791 488 × 2 = 1 + 0.999 999 994 691 231 443 582 976;
  • 47) 0.999 999 994 691 231 443 582 976 × 2 = 1 + 0.999 999 989 382 462 887 165 952;
  • 48) 0.999 999 989 382 462 887 165 952 × 2 = 1 + 0.999 999 978 764 925 774 331 904;
  • 49) 0.999 999 978 764 925 774 331 904 × 2 = 1 + 0.999 999 957 529 851 548 663 808;
  • 50) 0.999 999 957 529 851 548 663 808 × 2 = 1 + 0.999 999 915 059 703 097 327 616;
  • 51) 0.999 999 915 059 703 097 327 616 × 2 = 1 + 0.999 999 830 119 406 194 655 232;
  • 52) 0.999 999 830 119 406 194 655 232 × 2 = 1 + 0.999 999 660 238 812 389 310 464;
  • 53) 0.999 999 660 238 812 389 310 464 × 2 = 1 + 0.999 999 320 477 624 778 620 928;
  • 54) 0.999 999 320 477 624 778 620 928 × 2 = 1 + 0.999 998 640 955 249 557 241 856;

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:


0.000 000 000 742 147 676 646 634(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 634(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:


0.000 000 000 742 147 676 646 634(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 -0.000 000 000 742 147 676 646 634 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