-0.000 000 000 742 147 676 633 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 633 4(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
-0.000 000 000 742 147 676 633 4(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:

|-0.000 000 000 742 147 676 633 4| = 0.000 000 000 742 147 676 633 4


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: 0.000 000 000 742 147 676 633 4.

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) 0.000 000 000 742 147 676 633 4 × 2 = 0 + 0.000 000 001 484 295 353 266 8;
  • 2) 0.000 000 001 484 295 353 266 8 × 2 = 0 + 0.000 000 002 968 590 706 533 6;
  • 3) 0.000 000 002 968 590 706 533 6 × 2 = 0 + 0.000 000 005 937 181 413 067 2;
  • 4) 0.000 000 005 937 181 413 067 2 × 2 = 0 + 0.000 000 011 874 362 826 134 4;
  • 5) 0.000 000 011 874 362 826 134 4 × 2 = 0 + 0.000 000 023 748 725 652 268 8;
  • 6) 0.000 000 023 748 725 652 268 8 × 2 = 0 + 0.000 000 047 497 451 304 537 6;
  • 7) 0.000 000 047 497 451 304 537 6 × 2 = 0 + 0.000 000 094 994 902 609 075 2;
  • 8) 0.000 000 094 994 902 609 075 2 × 2 = 0 + 0.000 000 189 989 805 218 150 4;
  • 9) 0.000 000 189 989 805 218 150 4 × 2 = 0 + 0.000 000 379 979 610 436 300 8;
  • 10) 0.000 000 379 979 610 436 300 8 × 2 = 0 + 0.000 000 759 959 220 872 601 6;
  • 11) 0.000 000 759 959 220 872 601 6 × 2 = 0 + 0.000 001 519 918 441 745 203 2;
  • 12) 0.000 001 519 918 441 745 203 2 × 2 = 0 + 0.000 003 039 836 883 490 406 4;
  • 13) 0.000 003 039 836 883 490 406 4 × 2 = 0 + 0.000 006 079 673 766 980 812 8;
  • 14) 0.000 006 079 673 766 980 812 8 × 2 = 0 + 0.000 012 159 347 533 961 625 6;
  • 15) 0.000 012 159 347 533 961 625 6 × 2 = 0 + 0.000 024 318 695 067 923 251 2;
  • 16) 0.000 024 318 695 067 923 251 2 × 2 = 0 + 0.000 048 637 390 135 846 502 4;
  • 17) 0.000 048 637 390 135 846 502 4 × 2 = 0 + 0.000 097 274 780 271 693 004 8;
  • 18) 0.000 097 274 780 271 693 004 8 × 2 = 0 + 0.000 194 549 560 543 386 009 6;
  • 19) 0.000 194 549 560 543 386 009 6 × 2 = 0 + 0.000 389 099 121 086 772 019 2;
  • 20) 0.000 389 099 121 086 772 019 2 × 2 = 0 + 0.000 778 198 242 173 544 038 4;
  • 21) 0.000 778 198 242 173 544 038 4 × 2 = 0 + 0.001 556 396 484 347 088 076 8;
  • 22) 0.001 556 396 484 347 088 076 8 × 2 = 0 + 0.003 112 792 968 694 176 153 6;
  • 23) 0.003 112 792 968 694 176 153 6 × 2 = 0 + 0.006 225 585 937 388 352 307 2;
  • 24) 0.006 225 585 937 388 352 307 2 × 2 = 0 + 0.012 451 171 874 776 704 614 4;
  • 25) 0.012 451 171 874 776 704 614 4 × 2 = 0 + 0.024 902 343 749 553 409 228 8;
  • 26) 0.024 902 343 749 553 409 228 8 × 2 = 0 + 0.049 804 687 499 106 818 457 6;
  • 27) 0.049 804 687 499 106 818 457 6 × 2 = 0 + 0.099 609 374 998 213 636 915 2;
  • 28) 0.099 609 374 998 213 636 915 2 × 2 = 0 + 0.199 218 749 996 427 273 830 4;
  • 29) 0.199 218 749 996 427 273 830 4 × 2 = 0 + 0.398 437 499 992 854 547 660 8;
  • 30) 0.398 437 499 992 854 547 660 8 × 2 = 0 + 0.796 874 999 985 709 095 321 6;
  • 31) 0.796 874 999 985 709 095 321 6 × 2 = 1 + 0.593 749 999 971 418 190 643 2;
  • 32) 0.593 749 999 971 418 190 643 2 × 2 = 1 + 0.187 499 999 942 836 381 286 4;
  • 33) 0.187 499 999 942 836 381 286 4 × 2 = 0 + 0.374 999 999 885 672 762 572 8;
  • 34) 0.374 999 999 885 672 762 572 8 × 2 = 0 + 0.749 999 999 771 345 525 145 6;
  • 35) 0.749 999 999 771 345 525 145 6 × 2 = 1 + 0.499 999 999 542 691 050 291 2;
  • 36) 0.499 999 999 542 691 050 291 2 × 2 = 0 + 0.999 999 999 085 382 100 582 4;
  • 37) 0.999 999 999 085 382 100 582 4 × 2 = 1 + 0.999 999 998 170 764 201 164 8;
  • 38) 0.999 999 998 170 764 201 164 8 × 2 = 1 + 0.999 999 996 341 528 402 329 6;
  • 39) 0.999 999 996 341 528 402 329 6 × 2 = 1 + 0.999 999 992 683 056 804 659 2;
  • 40) 0.999 999 992 683 056 804 659 2 × 2 = 1 + 0.999 999 985 366 113 609 318 4;
  • 41) 0.999 999 985 366 113 609 318 4 × 2 = 1 + 0.999 999 970 732 227 218 636 8;
  • 42) 0.999 999 970 732 227 218 636 8 × 2 = 1 + 0.999 999 941 464 454 437 273 6;
  • 43) 0.999 999 941 464 454 437 273 6 × 2 = 1 + 0.999 999 882 928 908 874 547 2;
  • 44) 0.999 999 882 928 908 874 547 2 × 2 = 1 + 0.999 999 765 857 817 749 094 4;
  • 45) 0.999 999 765 857 817 749 094 4 × 2 = 1 + 0.999 999 531 715 635 498 188 8;
  • 46) 0.999 999 531 715 635 498 188 8 × 2 = 1 + 0.999 999 063 431 270 996 377 6;
  • 47) 0.999 999 063 431 270 996 377 6 × 2 = 1 + 0.999 998 126 862 541 992 755 2;
  • 48) 0.999 998 126 862 541 992 755 2 × 2 = 1 + 0.999 996 253 725 083 985 510 4;
  • 49) 0.999 996 253 725 083 985 510 4 × 2 = 1 + 0.999 992 507 450 167 971 020 8;
  • 50) 0.999 992 507 450 167 971 020 8 × 2 = 1 + 0.999 985 014 900 335 942 041 6;
  • 51) 0.999 985 014 900 335 942 041 6 × 2 = 1 + 0.999 970 029 800 671 884 083 2;
  • 52) 0.999 970 029 800 671 884 083 2 × 2 = 1 + 0.999 940 059 601 343 768 166 4;
  • 53) 0.999 940 059 601 343 768 166 4 × 2 = 1 + 0.999 880 119 202 687 536 332 8;
  • 54) 0.999 880 119 202 687 536 332 8 × 2 = 1 + 0.999 760 238 405 375 072 665 6;

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:


0.000 000 000 742 147 676 633 4(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 633 4(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:


0.000 000 000 742 147 676 633 4(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 -0.000 000 000 742 147 676 633 4 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