-0.000 000 000 742 147 676 626 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 626 2(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
-0.000 000 000 742 147 676 626 2(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:

|-0.000 000 000 742 147 676 626 2| = 0.000 000 000 742 147 676 626 2


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: 0.000 000 000 742 147 676 626 2.

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) 0.000 000 000 742 147 676 626 2 × 2 = 0 + 0.000 000 001 484 295 353 252 4;
  • 2) 0.000 000 001 484 295 353 252 4 × 2 = 0 + 0.000 000 002 968 590 706 504 8;
  • 3) 0.000 000 002 968 590 706 504 8 × 2 = 0 + 0.000 000 005 937 181 413 009 6;
  • 4) 0.000 000 005 937 181 413 009 6 × 2 = 0 + 0.000 000 011 874 362 826 019 2;
  • 5) 0.000 000 011 874 362 826 019 2 × 2 = 0 + 0.000 000 023 748 725 652 038 4;
  • 6) 0.000 000 023 748 725 652 038 4 × 2 = 0 + 0.000 000 047 497 451 304 076 8;
  • 7) 0.000 000 047 497 451 304 076 8 × 2 = 0 + 0.000 000 094 994 902 608 153 6;
  • 8) 0.000 000 094 994 902 608 153 6 × 2 = 0 + 0.000 000 189 989 805 216 307 2;
  • 9) 0.000 000 189 989 805 216 307 2 × 2 = 0 + 0.000 000 379 979 610 432 614 4;
  • 10) 0.000 000 379 979 610 432 614 4 × 2 = 0 + 0.000 000 759 959 220 865 228 8;
  • 11) 0.000 000 759 959 220 865 228 8 × 2 = 0 + 0.000 001 519 918 441 730 457 6;
  • 12) 0.000 001 519 918 441 730 457 6 × 2 = 0 + 0.000 003 039 836 883 460 915 2;
  • 13) 0.000 003 039 836 883 460 915 2 × 2 = 0 + 0.000 006 079 673 766 921 830 4;
  • 14) 0.000 006 079 673 766 921 830 4 × 2 = 0 + 0.000 012 159 347 533 843 660 8;
  • 15) 0.000 012 159 347 533 843 660 8 × 2 = 0 + 0.000 024 318 695 067 687 321 6;
  • 16) 0.000 024 318 695 067 687 321 6 × 2 = 0 + 0.000 048 637 390 135 374 643 2;
  • 17) 0.000 048 637 390 135 374 643 2 × 2 = 0 + 0.000 097 274 780 270 749 286 4;
  • 18) 0.000 097 274 780 270 749 286 4 × 2 = 0 + 0.000 194 549 560 541 498 572 8;
  • 19) 0.000 194 549 560 541 498 572 8 × 2 = 0 + 0.000 389 099 121 082 997 145 6;
  • 20) 0.000 389 099 121 082 997 145 6 × 2 = 0 + 0.000 778 198 242 165 994 291 2;
  • 21) 0.000 778 198 242 165 994 291 2 × 2 = 0 + 0.001 556 396 484 331 988 582 4;
  • 22) 0.001 556 396 484 331 988 582 4 × 2 = 0 + 0.003 112 792 968 663 977 164 8;
  • 23) 0.003 112 792 968 663 977 164 8 × 2 = 0 + 0.006 225 585 937 327 954 329 6;
  • 24) 0.006 225 585 937 327 954 329 6 × 2 = 0 + 0.012 451 171 874 655 908 659 2;
  • 25) 0.012 451 171 874 655 908 659 2 × 2 = 0 + 0.024 902 343 749 311 817 318 4;
  • 26) 0.024 902 343 749 311 817 318 4 × 2 = 0 + 0.049 804 687 498 623 634 636 8;
  • 27) 0.049 804 687 498 623 634 636 8 × 2 = 0 + 0.099 609 374 997 247 269 273 6;
  • 28) 0.099 609 374 997 247 269 273 6 × 2 = 0 + 0.199 218 749 994 494 538 547 2;
  • 29) 0.199 218 749 994 494 538 547 2 × 2 = 0 + 0.398 437 499 988 989 077 094 4;
  • 30) 0.398 437 499 988 989 077 094 4 × 2 = 0 + 0.796 874 999 977 978 154 188 8;
  • 31) 0.796 874 999 977 978 154 188 8 × 2 = 1 + 0.593 749 999 955 956 308 377 6;
  • 32) 0.593 749 999 955 956 308 377 6 × 2 = 1 + 0.187 499 999 911 912 616 755 2;
  • 33) 0.187 499 999 911 912 616 755 2 × 2 = 0 + 0.374 999 999 823 825 233 510 4;
  • 34) 0.374 999 999 823 825 233 510 4 × 2 = 0 + 0.749 999 999 647 650 467 020 8;
  • 35) 0.749 999 999 647 650 467 020 8 × 2 = 1 + 0.499 999 999 295 300 934 041 6;
  • 36) 0.499 999 999 295 300 934 041 6 × 2 = 0 + 0.999 999 998 590 601 868 083 2;
  • 37) 0.999 999 998 590 601 868 083 2 × 2 = 1 + 0.999 999 997 181 203 736 166 4;
  • 38) 0.999 999 997 181 203 736 166 4 × 2 = 1 + 0.999 999 994 362 407 472 332 8;
  • 39) 0.999 999 994 362 407 472 332 8 × 2 = 1 + 0.999 999 988 724 814 944 665 6;
  • 40) 0.999 999 988 724 814 944 665 6 × 2 = 1 + 0.999 999 977 449 629 889 331 2;
  • 41) 0.999 999 977 449 629 889 331 2 × 2 = 1 + 0.999 999 954 899 259 778 662 4;
  • 42) 0.999 999 954 899 259 778 662 4 × 2 = 1 + 0.999 999 909 798 519 557 324 8;
  • 43) 0.999 999 909 798 519 557 324 8 × 2 = 1 + 0.999 999 819 597 039 114 649 6;
  • 44) 0.999 999 819 597 039 114 649 6 × 2 = 1 + 0.999 999 639 194 078 229 299 2;
  • 45) 0.999 999 639 194 078 229 299 2 × 2 = 1 + 0.999 999 278 388 156 458 598 4;
  • 46) 0.999 999 278 388 156 458 598 4 × 2 = 1 + 0.999 998 556 776 312 917 196 8;
  • 47) 0.999 998 556 776 312 917 196 8 × 2 = 1 + 0.999 997 113 552 625 834 393 6;
  • 48) 0.999 997 113 552 625 834 393 6 × 2 = 1 + 0.999 994 227 105 251 668 787 2;
  • 49) 0.999 994 227 105 251 668 787 2 × 2 = 1 + 0.999 988 454 210 503 337 574 4;
  • 50) 0.999 988 454 210 503 337 574 4 × 2 = 1 + 0.999 976 908 421 006 675 148 8;
  • 51) 0.999 976 908 421 006 675 148 8 × 2 = 1 + 0.999 953 816 842 013 350 297 6;
  • 52) 0.999 953 816 842 013 350 297 6 × 2 = 1 + 0.999 907 633 684 026 700 595 2;
  • 53) 0.999 907 633 684 026 700 595 2 × 2 = 1 + 0.999 815 267 368 053 401 190 4;
  • 54) 0.999 815 267 368 053 401 190 4 × 2 = 1 + 0.999 630 534 736 106 802 380 8;

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:


0.000 000 000 742 147 676 626 2(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 626 2(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:


0.000 000 000 742 147 676 626 2(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 -0.000 000 000 742 147 676 626 2 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