-4.123 105 625 617 660 549 821 409 855 974 077 025 147 199 225 277 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -4.123 105 625 617 660 549 821 409 855 974 077 025 147 199 225 277(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
-4.123 105 625 617 660 549 821 409 855 974 077 025 147 199 225 277(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:

|-4.123 105 625 617 660 549 821 409 855 974 077 025 147 199 225 277| = 4.123 105 625 617 660 549 821 409 855 974 077 025 147 199 225 277


2. First, convert to binary (in base 2) the integer part: 4.
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;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

4(10) =

100(2)


4. Convert to binary (base 2) the fractional part: 0.123 105 625 617 660 549 821 409 855 974 077 025 147 199 225 277.

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) 0.123 105 625 617 660 549 821 409 855 974 077 025 147 199 225 277 × 2 = 0 + 0.246 211 251 235 321 099 642 819 711 948 154 050 294 398 450 554;
  • 2) 0.246 211 251 235 321 099 642 819 711 948 154 050 294 398 450 554 × 2 = 0 + 0.492 422 502 470 642 199 285 639 423 896 308 100 588 796 901 108;
  • 3) 0.492 422 502 470 642 199 285 639 423 896 308 100 588 796 901 108 × 2 = 0 + 0.984 845 004 941 284 398 571 278 847 792 616 201 177 593 802 216;
  • 4) 0.984 845 004 941 284 398 571 278 847 792 616 201 177 593 802 216 × 2 = 1 + 0.969 690 009 882 568 797 142 557 695 585 232 402 355 187 604 432;
  • 5) 0.969 690 009 882 568 797 142 557 695 585 232 402 355 187 604 432 × 2 = 1 + 0.939 380 019 765 137 594 285 115 391 170 464 804 710 375 208 864;
  • 6) 0.939 380 019 765 137 594 285 115 391 170 464 804 710 375 208 864 × 2 = 1 + 0.878 760 039 530 275 188 570 230 782 340 929 609 420 750 417 728;
  • 7) 0.878 760 039 530 275 188 570 230 782 340 929 609 420 750 417 728 × 2 = 1 + 0.757 520 079 060 550 377 140 461 564 681 859 218 841 500 835 456;
  • 8) 0.757 520 079 060 550 377 140 461 564 681 859 218 841 500 835 456 × 2 = 1 + 0.515 040 158 121 100 754 280 923 129 363 718 437 683 001 670 912;
  • 9) 0.515 040 158 121 100 754 280 923 129 363 718 437 683 001 670 912 × 2 = 1 + 0.030 080 316 242 201 508 561 846 258 727 436 875 366 003 341 824;
  • 10) 0.030 080 316 242 201 508 561 846 258 727 436 875 366 003 341 824 × 2 = 0 + 0.060 160 632 484 403 017 123 692 517 454 873 750 732 006 683 648;
  • 11) 0.060 160 632 484 403 017 123 692 517 454 873 750 732 006 683 648 × 2 = 0 + 0.120 321 264 968 806 034 247 385 034 909 747 501 464 013 367 296;
  • 12) 0.120 321 264 968 806 034 247 385 034 909 747 501 464 013 367 296 × 2 = 0 + 0.240 642 529 937 612 068 494 770 069 819 495 002 928 026 734 592;
  • 13) 0.240 642 529 937 612 068 494 770 069 819 495 002 928 026 734 592 × 2 = 0 + 0.481 285 059 875 224 136 989 540 139 638 990 005 856 053 469 184;
  • 14) 0.481 285 059 875 224 136 989 540 139 638 990 005 856 053 469 184 × 2 = 0 + 0.962 570 119 750 448 273 979 080 279 277 980 011 712 106 938 368;
  • 15) 0.962 570 119 750 448 273 979 080 279 277 980 011 712 106 938 368 × 2 = 1 + 0.925 140 239 500 896 547 958 160 558 555 960 023 424 213 876 736;
  • 16) 0.925 140 239 500 896 547 958 160 558 555 960 023 424 213 876 736 × 2 = 1 + 0.850 280 479 001 793 095 916 321 117 111 920 046 848 427 753 472;
  • 17) 0.850 280 479 001 793 095 916 321 117 111 920 046 848 427 753 472 × 2 = 1 + 0.700 560 958 003 586 191 832 642 234 223 840 093 696 855 506 944;
  • 18) 0.700 560 958 003 586 191 832 642 234 223 840 093 696 855 506 944 × 2 = 1 + 0.401 121 916 007 172 383 665 284 468 447 680 187 393 711 013 888;
  • 19) 0.401 121 916 007 172 383 665 284 468 447 680 187 393 711 013 888 × 2 = 0 + 0.802 243 832 014 344 767 330 568 936 895 360 374 787 422 027 776;
  • 20) 0.802 243 832 014 344 767 330 568 936 895 360 374 787 422 027 776 × 2 = 1 + 0.604 487 664 028 689 534 661 137 873 790 720 749 574 844 055 552;
  • 21) 0.604 487 664 028 689 534 661 137 873 790 720 749 574 844 055 552 × 2 = 1 + 0.208 975 328 057 379 069 322 275 747 581 441 499 149 688 111 104;
  • 22) 0.208 975 328 057 379 069 322 275 747 581 441 499 149 688 111 104 × 2 = 0 + 0.417 950 656 114 758 138 644 551 495 162 882 998 299 376 222 208;
  • 23) 0.417 950 656 114 758 138 644 551 495 162 882 998 299 376 222 208 × 2 = 0 + 0.835 901 312 229 516 277 289 102 990 325 765 996 598 752 444 416;
  • 24) 0.835 901 312 229 516 277 289 102 990 325 765 996 598 752 444 416 × 2 = 1 + 0.671 802 624 459 032 554 578 205 980 651 531 993 197 504 888 832;

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:


0.123 105 625 617 660 549 821 409 855 974 077 025 147 199 225 277(10) =

0.0001 1111 1000 0011 1101 1001(2)

6. Positive number before normalization:

4.123 105 625 617 660 549 821 409 855 974 077 025 147 199 225 277(10) =

100.0001 1111 1000 0011 1101 1001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


4.123 105 625 617 660 549 821 409 855 974 077 025 147 199 225 277(10) =

100.0001 1111 1000 0011 1101 1001(2) =

100.0001 1111 1000 0011 1101 1001(2) × 20 =

1.0000 0111 1110 0000 1111 0110 01(2) × 22


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): 2

Mantissa (not normalized):
1.0000 0111 1110 0000 1111 0110 01


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

2 + 2(8-1) - 1 =

(2 + 127)(10) =

129(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;
  • 129 ÷ 2 = 64 + 1;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

129(10) =

1000 0001(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 000 0011 1111 0000 0111 1011 001 =

000 0011 1111 0000 0111 1011


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) =
1000 0001

Mantissa (23 bits) =
000 0011 1111 0000 0111 1011


Decimal number -4.123 105 625 617 660 549 821 409 855 974 077 025 147 199 225 277 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 1000 0001 - 000 0011 1111 0000 0111 1011

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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